2

1.1 Electric Charge

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Describe the concept of electric charge
Explain qualitatively the force electric charge creates

You are certainly familiar with electronic devices that you activate with the click of a switch, from computers to cell phones to television. And you have certainly seen electricity in a flash of lightning during a heavy thunderstorm. But you have also most likely experienced electrical effects in other ways, maybe without realizing that an electric force was involved. Let’s take a look at some of these activities and see what we can learn from them about electric charges and forces.

Discoveries

You have probably experienced the phenomenon of static electricity: When you first take clothes out of a dryer, many (not all) of them tend to stick together; for some fabrics, they can be very difficult to separate. Another example occurs if you take a woolen sweater off quickly—you can feel (and hear) the static electricity pulling on your clothes, and perhaps even your hair. If you comb your hair on a dry day and then put the comb close to a thin stream of water coming out of a faucet, you will find that the water stream bends toward (is attracted to) the comb (Figure 1.1.1).

(Figure 1.1.1) $\begin{gather*}.\end{gather*}$

A photograph of a stream of water bending sideways as it is attracted to a comb.

Figure 1.1.1. An electrically charged comb attracts a stream of water from a distance. Note that the water is not touching the comb. (credit: Jane Whitney)

Suppose you bring the comb close to some small strips of paper; the strips of paper are attracted to the comb and even cling to it (Figure 1.1.2 ). In the kitchen, quickly pull a length of plastic cling wrap off the roll; it will tend to cling to most any nonmetallic material (such as plastic, glass, or food). If you rub a balloon on a wall for a few seconds, it will stick to the wall. Probably the most annoying effect of static electricity is getting shocked by a doorknob (or a friend) after shuffling your feet on some types of carpeting.

(Figure 1.1.2) $\begin{gather*}.\end{gather*}$

A photograph of thin strips of paper stuck to a plastic comb.

Figure 1.1.2. After being used to comb hair, this comb attracts small strips of paper from a distance, without physical contact. Investigation of this behavior helped lead to the concept of the electric force.

Many of these phenomena have been known for centuries. The ancient Greek philosopher Thales of Miletus (624–546 BCE) recorded that when amber (a hard, translucent, fossilized resin from extinct trees) was vigorously rubbed with a piece of fur, a force was created that caused the fur and the amber to be attracted to each other (Figure 1.1.3 ). Additionally, he found that the rubbed amber would not only attract the fur, and the fur attract the amber, but they both could affect other (nonmetallic) objects, even if not in contact with those objects (Figure 1.1.4).

(Figure 1.1.3) $\begin{gather*}.\end{gather*}$

A photograph of a piece of gold-colored amber from Malaysia that has been rubbed and polished to a smooth, rounded shape.

Figure 1.1.3. Borneo amber is mined in Sabah, Malaysia, from shale-sandstone-mudstone veins. When a piece of amber is rubbed with a piece of fur, the amber gains more electrons, giving it a net negative charge. At the same time, the fur, having lost electrons, becomes positively charged. (credit: “Sebakoamber”/Wikimedia Commons)

(Figure 1.1.4) $\begin{gather*}.\end{gather*}$

Figure a shows a piece of amber and a piece of cloth. The amber has two negative charges and two positive charges, while the cloth has three of each. In figure B, two arrows are shown going through the amber, and another two arrows coming out of the amber. In figure C, the amber now has two positive charges and four negative charges, while the cloth has three positive charges and only one remaining negative charge.

Figure 1.1.4. When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity for electrons than another. (a) Both the amber and cloth are originally neutral, with equal positive and negative charges. Only a tiny fraction of the charges are involved, and only a few of them are shown here. (b) When rubbed together, some negative charge is transferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have net charges, but the absolute value of the net positive and negative charges will be equal.

The English physicist William Gilbert (1544–1603) also studied this attractive force, using various substances. He worked with amber, and, in addition, he experimented with rock crystal and various precious and semi-precious gemstones. He also experimented with several metals. He found that the metals never exhibited this force, whereas the minerals did. Moreover, although an electrified amber rod would attract a piece of fur, it would repel another electrified amber rod; similarly, two electrified pieces of fur would repel each other.

This suggested there were two types of an electric property; this property eventually came to be called electric charge. The difference between the two types of electric charge is in the directions of the electric forces that each type of charge causes: These forces are repulsive when the same type of charge exists on two interacting objects and attractive when the charges are of opposite types. The SI unit of electric charge is the coulomb (C), after the French physicist Charles Augustine de Coulomb (1736–1806).

The most peculiar aspect of this new force is that it does not require physical contact between the two objects in order to cause an acceleration. This is an example of a so-called “long-range” force. (Or, as Albert Einstein later phrased it, “action at a distance.”) With the exception of gravity, all other forces we have discussed so far act only when the two interacting objects actually touch.

The American physicist and statesman Benjamin Franklin found that he could concentrate charge in a “Leyden jar,” which was essentially a glass jar with two sheets of metal foil, one inside and one outside, with the glass between them (Figure 1.1.5). This created a large electric force between the two foil sheets.

(Figure 1.1.5) $\begin{gather*}.\end{gather*}$

This figure is an illustration of a Leyden Jar. A layer of tin foil is wrapped around the outside and the inside surfaces of a glass jar. A wire is attached to the inner foil and connected to a metal rod that extends out through a stopper at the top of the jar. The inner foil is marked as having positive charge and the outer as having negative charge.

Figure 1.1.5. A Leyden jar (an early version of what is now called a capacitor) allowed experimenters to store large amounts of electric charge. Benjamin Franklin used such a jar to demonstrate that lightning behaved exactly like the electricity he got from the equipment in his laboratory.

Franklin pointed out that the observed behavior could be explained by supposing that one of the two types of charge remained motionless, while the other type of charge flowed from one piece of foil to the other. He further suggested that an excess of what he called this “electrical fluid” be called “positive electricity” and the deficiency of it be called “negative electricity.” His suggestion, with some minor modifications, is the model we use today. (With the experiments that he was able to do, this was a pure guess; he had no way of actually determining the sign of the moving charge. Unfortunately, he guessed wrong; we now know that the charges that flow are the ones Franklin labeled negative, and the positive charges remain largely motionless. Fortunately, as we’ll see, it makes no practical or theoretical difference which choice we make, as long as we stay consistent with our choice.)

Let’s list the specific observations that we have of this electric force:

The force acts without physical contact between the two objects.
The force can be either attractive or repulsive: If two interacting objects carry the same sign of charge, the force is repulsive; if the charges are of opposite sign, the force is attractive. These interactions are referred to as electrostatic repulsion and electrostatic attraction, respectively.
Not all objects are affected by this force.
The magnitude of the force decreases (rapidly) with increasing separation distance between the objects.

To be more precise, we find experimentally that the magnitude of the force decreases as the square of the distance between the two interacting objects increases. Thus, for example, when the distance between two interacting objects is doubled, the force between them decreases to one fourth what it was in the original system. We can also observe that the surroundings of the charged objects affect the magnitude of the force. However, we will explore this issue in a later chapter.

Properties of Electric Charge

In addition to the existence of two types of charge, several other properties of charge have been discovered.

Charge is quantized. This means that electric charge comes in discrete amounts, and there is a smallest possible amount of charge that an object can have. In the SI system, this smallest amount is $e\equiv1.602\times10^{-19}$ . No free particle can have less charge than this, and, therefore, the charge on any object—the charge on all objects—must be an integer multiple of this amount. All macroscopic, charged objects have charge because electrons have either been added or taken away from them, resulting in a net charge.
The magnitude of the charge is independent of the type. Phrased another way, the smallest possible positive charge (to four significant figures) is $+1.602\times10^{-19}~\mathrm{C}$ , and the smallest possible negative charge is $-1.602\times10^{-19}~\mathrm{C}$ ; these values are exactly equal. This is simply how the laws of physics in our universe turned out.
Charge is conserved. Charge can neither be created nor destroyed; it can only be transferred from place to place, from one object to another. Frequently, we speak of two charges “canceling”; this is verbal shorthand. It means that if two objects that have equal and opposite charges are physically close to each other, then the (oppositely directed) forces they apply on some other charged object cancel, for a net force of zero. It is important that you understand that the charges on the objects by no means disappear, however. The net charge of the universe is constant.
Charge is conserved in closed systems. In principle, if a negative charge disappeared from your lab bench and reappeared on the Moon, conservation of charge would still hold. However, this never happens. If the total charge you have in your local system on your lab bench is changing, there will be a measurable flow of charge into or out of the system. Again, charges can and do move around, and their effects can and do cancel, but the net charge in your local environment (if closed) is conserved. The last two items are both referred to as the law of conservation of charge.

The Source of Charges: The Structure of the Atom

Once it became clear that all matter was composed of particles that came to be called atoms, it also quickly became clear that the constituents of the atom included both positively charged particles and negatively charged particles. The next question was, what are the physical properties of those electrically charged particles?

The negatively charged particle was the first one to be discovered. In 1897, the English physicist J. J. Thomson was studying what was then known as cathode rays. Some years before, the English physicist William Crookes had shown that these “rays” were negatively charged, but his experiments were unable to tell any more than that. (The fact that they carried a negative electric charge was strong evidence that these were not rays at all, but particles.) Thomson prepared a pure beam of these particles and sent them through crossed electric and magnetic fields, and adjusted the various field strengths until the net deflection of the beam was zero. With this experiment, he was able to determine the charge-to-mass ratio of the particle. This ratio showed that the mass of the particle was much smaller than that of any other previously known particle—1837 times smaller, in fact. Eventually, this particle came to be called the electron.

Since the atom as a whole is electrically neutral, the next question was to determine how the positive and negative charges are distributed within the atom. Thomson himself imagined that his electrons were embedded within a sort of positively charged paste, smeared out throughout the volume of the atom. However, in 1908, the New Zealand physicist Ernest Rutherford showed that the positive charges of the atom existed within a tiny core—called a nucleus—that took up only a very tiny fraction of the overall volume of the atom, but held over $99\%$ of the mass. In addition, he showed that the negatively charged electrons perpetually orbited about this nucleus, forming a sort of electrically charged cloud that surrounds the nucleus (Figure 1.1.6). Rutherford concluded that the nucleus was constructed of small, massive particles that he named protons.

(Figure 1.1.6) $\begin{gather*}.\end{gather*}$

An illustration of the simplified model of a hydrogen atom. The nucleus is shown as a small dark, solid sphere at he center of an electron cloud.

Figure 1.1.6. This simplified model of a hydrogen atom shows a positively charged nucleus (consisting, in the case of hydrogen, of a single proton), surrounded by an electron “cloud.” The charge of the electron cloud is equal (and opposite in sign) to the charge of the nucleus, but the electron does not have a definite location in space; hence, its representation here is as a cloud. Normal macroscopic amounts of matter contain immense numbers of atoms and molecules, and, hence, even greater numbers of individual negative and positive charges.

Since it was known that different atoms have different masses, and that ordinarily atoms are electrically neutral, it was natural to suppose that different atoms have different numbers of protons in their nucleus, with an equal number of negatively charged electrons orbiting about the positively charged nucleus, thus making the atoms overall electrically neutral. However, it was soon discovered that although the lightest atom, hydrogen, did indeed have a single proton as its nucleus, the next heaviest atom—helium—has twice the number of protons (two), but four times the mass of hydrogen.

This mystery was resolved in 1932 by the English physicist James Chadwick, with the discovery of the neutron. The neutron is, essentially, an electrically neutral twin of the proton, with no electric charge, but (nearly) identical mass to the proton. The helium nucleus therefore has two neutrons along with its two protons. (Later experiments were to show that although the neutron is electrically neutral overall, it does have an internal charge structure. Furthermore, although the masses of the neutron and the proton are nearly equal, they aren’t exactly equal: The neutron’s mass is very slightly larger than the mass of the proton. That slight mass excess turned out to be of great importance.)

Thus, in 1932, the picture of the atom was of a small, massive nucleus constructed of a combination of protons and neutrons, surrounded by a collection of electrons whose combined motion formed a sort of negatively charged “cloud” around the nucleus (Figure 1.1.7). In an electrically neutral atom, the total negative charge of the collection of electrons is equal to the total positive charge in the nucleus. The very low-mass electrons can be more or less easily removed or added to an atom, changing the net charge on the atom (though without changing its type). An atom that has had the charge altered in this way is called an ion. Positive ions have had electrons removed, whereas negative ions have had excess electrons added. We also use this term to describe molecules that are not electrically neutral.

(Figure 1.1.7) $\begin{gather*}.\end{gather*}$

An illustration of the simplified model of a carbon atom. The nucleus is shown as a cluster of small blue and red spheres. The blue spheres represent neutrons and the red ones represent protons. The nucleus is surrounded by an electron cloud, represented by a shaded blue region with six darker spots representing the six localized electrons.

Figure 1.1.7. The nucleus of a carbon atom is composed of six protons and six neutrons. As in hydrogen, the surrounding six electrons do not have definite locations and so can be considered to be a sort of cloud surrounding the nucleus.

The story of the atom does not stop there, however. In the latter part of the twentieth century, many more subatomic particles were discovered in the nucleus of the atom: pions, neutrinos, and quarks, among others. With the exception of the photon, none of these particles are directly relevant to the study of electromagnetism, so we will not discuss them further in this course.

A Note on Terminology

As noted previously, electric charge is a property that an object can have. This is similar to how an object can have a property that we call mass, a property that we call density, a property that we call temperature, and so on. Technically, we should always say something like, “Suppose we have a particle that carries a charge of $3~\mu\mathrm{C}$ ” However, it is very common to say instead, “Suppose we have a $3$ – $\mu\mathrm{C}$ charge.” Similarly, we often say something like, “Six charges are located at the vertices of a regular hexagon.” A charge is not a particle; rather, it is a property of a particle. Nevertheless, this terminology is extremely common (and is frequently used in this book, as it is everywhere else). So, keep in the back of your mind what we really mean when we refer to a “charge.

3

1.2 Conductors, Insulators, and Charging by Induction

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Explain what a conductor is
Explain what an insulator is
List the differences and similarities between conductors and insulators
Describe the process of charging by induction

In the preceding section, we said that scientists were able to create electric charge only on nonmetallic materials and never on metals. To understand why this is the case, you have to understand more about the nature and structure of atoms. In this section, we discuss how and why electric charges do—or do not—move through materials (Figure 1.2.1). A more complete description is given in a later chapter.

(Figure 1.2.1) $\begin{gather*}.\end{gather*}$

A photograph of a black power charging unit that connects a laptop to an electrical outlet, allowing the laptop to be charged.

Figure 1.2.1. This power adapter uses metal wires and connectors to conduct electricity from the wall socket to a laptop computer. The conducting wires allow electrons to move freely through the cables, which are shielded by rubber and plastic. These materials act as insulators that don’t allow electric charge to escape outward. (credit: modification of work by “Evan-Amos”/Wikimedia Commons)

Conductors and Insulators

As discussed in the previous section, electrons surround the tiny nucleus in the form of a (comparatively) vast cloud of negative charge. However, this cloud does have a definite structure to it. Let’s consider an atom of the most commonly used conductor, copper.

There is an outermost electron that is only loosely bound to the atom’s nucleus. It can be easily dislodged; it then moves to a neighboring atom. In a large mass of copper atoms (such as a copper wire or a sheet of copper), these vast numbers of outermost electrons (one per atom) wander from atom to atom, and are the electrons that do the moving when electricity flows. These wandering, or “free,” electrons are called conduction electrons, and copper is therefore an excellent conductor (of electric charge). All conducting elements have a similar arrangement of their electrons, with one or two conduction electrons. This includes most metals.

Insulators, in contrast, are made from materials that lack conduction electrons; charge flows only with great difficulty, if at all. Even if excess charge is added to an insulating material, it cannot move, remaining indefinitely in place. This is why insulating materials exhibit the electrical attraction and repulsion forces described earlier, whereas conductors do not; any excess charge placed on a conductor would instantly flow away (due to mutual repulsion from existing charges), leaving no excess charge around to create forces. Charge cannot flow along or through an insulator, so its electric forces remain for long periods of time. (Charge will dissipate from an insulator, given enough time.) As it happens, amber, fur, and most semi-precious gems are insulators, as are materials like wood, glass, and plastic.

Charging by Induction

Let’s examine in more detail what happens in a conductor when an electrically charged object is brought close to it. As mentioned, the conduction electrons in the conductor are able to move with nearly complete freedom. As a result, when a charged insulator (such as a positively charged glass rod) is brought close to the conductor, the (total) charge on the insulator exerts an electric force on the conduction electrons. Since the rod is positively charged, the conduction electrons (which themselves are negatively charged) are attracted, flowing toward the insulator to the near side of the conductor (Figure 1.2.2).

Now, the conductor is still overall electrically neutral; the conduction electrons have changed position, but they are still in the conducting material. However, the conductor now has a charge distribution; the near end (the portion of the conductor closest to the insulator) now has more negative charge than positive charge, and the reverse is true of the end farthest from the insulator. The relocation of negative charges to the near side of the conductor results in an overall positive charge in the part of the conductor farthest from the insulator. We have thus created an electric charge distribution where one did not exist before. This process is referred to as inducing polarization—in this case, polarizing the conductor. The resulting separation of positive and negative charge is called polarization, and a material, or even a molecule, that exhibits polarization is said to be polarized. A similar situation occurs with a negatively charged insulator, but the resulting polarization is in the opposite direction.

(Figure 1.2.2) $\begin{gather*}.\end{gather*}$

A microscopic view of polarization is shown. A positively charged glass rod with positive signs is close to a neutral conducting sphere with a charge distribution. The negative charges on the sphere are on the side near the rod and positive charges are on the side opposite from the rod.

Figure 1.2.2. Induced polarization. A positively charged glass rod is brought near the left side of the conducting sphere, attracting negative charge and leaving the other side of the sphere positively charged. Although the sphere is overall still electrically neutral, it now has a charge distribution, so it can exert an electric force on other nearby charges. Furthermore, the distribution is such that it will be attracted to the glass rod.

The result is the formation of what is called an electric dipole, from a Latin phrase meaning “two ends.” The presence of electric charges on the insulator—and the electric forces they apply to the conduction electrons—creates, or “induces,” the dipole in the conductor.

Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, for example. If you run a plastic comb through your hair, the charged comb can pick up neutral pieces of paper. Figure 1.2.3 shows how the polarization of atoms and molecules in neutral objects results in their attraction to a charged object.

(Figure 1.2.3) $\begin{gather*}.\end{gather*}$

Microscopic views of objects are shown. In part a, a positive rod with positive signs is close to an insulator. The negative ends of all the molecules of the insulator are aligned toward the rod and positive ends of all molecules shown as spheres are away from the rod. In part b, a negative rod with negative signs is close to an insulator. The positive ends of all the molecules of the insulator are aligned toward the rod and negative ends of all molecules shown as spheres are away from the rod. In part c, a rod with negative signs is close to an insulator. Only the net charges are shown in the insulator. The insulator surface closer to the rod has positive signs. The other surface has negative signs.

Figure 1.2.3. Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive object brought near a neutral insulator polarizes its molecules. There is a slight shift in the distribution of the electrons orbiting the molecule, with unlike charges being brought nearer and like charges moved away. Since the electrostatic force decreases with distance, there is a net attraction. (b) A negative object produces the opposite polarization, but again attracts the neutral object. (c) The same effect occurs for a conductor; since the unlike charges are closer, there is a net attraction.

When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction. Thus, a positively charged glass rod attracts neutral pieces of paper, as will a negatively charged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separation of charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and show greater polarization effects than molecules with naturally uniform charge distributions.

When the two ends of a dipole can be separated, this method of charging by induction may be used to create charged objects without transferring charge. In Figure 1.2.4, we see two neutral metal spheres in contact with one another but insulated from the rest of the world. A positively charged rod is brought near one of them, attracting negative charge to that side, leaving the other sphere positively charged.

(Figure 1.2.4) $\begin{gather*}.\end{gather*}$

In part a, a pair of neutral metal spheres are in contact. In part b, a rod with positive charge is close to the surface of one of the sphere. Negative signs are shown on this surface near the rod and positive signs are shown on the farthest part of the surface of the other sphere. The charged rod causes separation of charge. In part c, the positively charged rod is near the spheres, and the spheres are not in contact. As in figure b, the outer surface of the sphere nearest the rod has negative signs and the far surface of the other sphere has positive signs. In part d, the glass rod is not shown. The charges are now on the inner surfaces of the metallic spheres. One sphere has negative signs and the other has positive signs facing each other.

Figure 1.2.4. Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with each other but insulated from the rest of the world. (b) A positively charged glass rod is brought near the sphere on the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres are separated before the rod is removed, thus separating negative and positive charges. (d) The spheres retain net charges after the inducing rod is removed—without ever having been touched by a charged object.

Another method of charging by induction is shown in Figure 1.2.5. The neutral metal sphere is polarized when a charged rod is brought near it. The sphere is then grounded, meaning that a conducting wire is run from the sphere to the ground. Since Earth is large and most of the ground is a good conductor, it can supply or accept excess charge easily. In this case, electrons are attracted to the sphere through a wire called the ground wire, because it supplies a conducting path to the ground. The ground connection is broken before the charged rod is removed, leaving the sphere with an excess charge opposite to that of the rod. Again, an opposite charge is achieved when charging by induction, and the charged rod loses none of its excess charge.

(Figure 1.2.5) $\begin{gather*}.\end{gather*}$

In part a, a rod with positive sign is brought near a neutral metal sphere. The surface of the sphere near the rod has negative signs and the surface far from it has positive signs. In part b, the sphere is connected to ground by a wire attached to the surface farthest from the rod. Negative charge is shown moving from the ground up to the sphere. The negative charges on the sphere near the rod are unaffected but there are fewer positive charges where the sphere is grounded. In part c, the sphere is disconnected from ground. The rod with positive sign is close to one surface of the sphere where negative charges are shown, and the other surface has no charges shown. In part d, the positive rod is absent, and the sphere has negative signs on it uniformly distributed on its surface.

Figure 1.2.5. Charging by induction using a ground connection. (a) A positively charged rod is brought near a neutral metal sphere, polarizing it. (b) The sphere is grounded, allowing electrons to be attracted from Earth’s ample supply. (c) The ground connection is broken. (d) The positive rod is removed, leaving the sphere with an induced negative charge.

4

1.3 Coulomb's Law

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Describe the electric force, both qualitatively and quantitatively
Calculate the force that charges exert on each other
Determine the direction of the electric force for different source charges
Correctly describe and apply the superposition principle for multiple source charges

Experiments with electric charges have shown that if two objects each have electric charge, then they exert an electric force on each other. The magnitude of the force is linearly proportional to the net charge on each object and inversely proportional to the square of the distance between them. (Interestingly, the force does not depend on the mass of the objects.) The direction of the force vector is along the imaginary line joining the two objects and is dictated by the signs of the charges involved.

Let

$q_1,~q_2=$ the net electric charges of the two objects;
$\vec{\mathbf{r}}_{12}=$ the vector displacement from $q_1$ to $q_2$ .

The electric force $\vec{\mathbf{F}}$ on one of the charges is proportional to the magnitude of its own charge and the magnitude of the other charge, and is inversely proportional to the square of the distance between them:

F\propto\frac{q_1q_2}{{r_{12}}^2}.

This proportionality becomes an equality with the introduction of a proportionality constant. For reasons that will become clear in a later chapter, the proportionality constant that we use is actually a collection of constants. (We discuss this constant shortly.)

COULOMB’S LAW

The electric force (or Coulomb force) between two electrically charged particles is equal to

(1.3.1) $\begin{equation*}\vec{\mathbf{F}}_{12}(r)=\frac{1}{4\pi\epsilon_0}\frac{|q_1q_2|}{{r_{12}}^2}\hat{\mathbf{r}}_{12}\end{equation*}$

We use absolute value signs around the product $q_1q_2$ because one of the charges may be negative, but the magnitude of the force is always positive. The unit vector $\hat{\mathbf{r}}$ points directly from the charge $q_1$ toward $q_2$ . If $q_1$ and $q_2$ have the same sign, the force vector on $q_2$ points away from $q_1$ ; if they have opposite signs, the force on $q_2$ points toward $q_1$ (Figure 1.3.1).

(Figure 1.3.1) $\begin{gather*}.\end{gather*}$

In part a, two charges q one and q two are shown separated by a distance r. Force vector arrow F one two points toward left and acts on q one. Force vector arrow F two one points toward right and acts on q two. Both forces act in opposite directions and are represented by arrows of same length. In part b, two charges q one and q two are shown at a distance r. Force vector arrow F one two points toward right and acts on q one. Force vector arrow F two one points toward left and acts on q two. Both forces act toward each other and are represented by arrows of same length.

Figure 1.3.1. The electrostatic force $\vec{\mathbf{F}}$ between point charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb’s law. Note that Newton’s third law (every force exerted creates an equal and opposite force) applies as usual—the force on $q_1$ is equal in magnitude and opposite in direction to the force it exerts on $q_2$ . (a) Like charges; (b) unlike charges.

It is important to note that the electric force is not constant; it is a function of the separation distance between the two charges. If either the test charge or the source charge (or both) move, then $\vec{\mathbf{r}}$ changes, and therefore so does the force. An immediate consequence of this is that direct application of Newton’s laws with this force can be mathematically difficult, depending on the specific problem at hand. It can (usually) be done, but we almost always look for easier methods of calculating whatever physical quantity we are interested in. (Conservation of energy is the most common choice.)

Finally, the new constant $\varepsilon_0$ in Coulomb’s law is called the permittivity of free space, or (better) the permittivity of vacuum. It has a very important physical meaning that we will discuss in a later chapter; for now, it is simply an empirical proportionality constant. Its numerical value (to three significant figures) turns out to be

$\varepsilon_0=8.85\times10^{-12}~\frac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2}.$

These units are required to give the force in Coulomb’s law the correct units of newtons. Note that in Coulomb’s law, the permittivity of vacuum is only part of the proportionality constant. For convenience, we often define a Coulomb’s constant:

$k_e=\frac{1}{4\pi\varepsilon_0}=8.99\times10^9~\frac{\mathrm{N}\cdot\mathrm{m}^2}{\mathrm{C}^2}.$

EXAMPLE 1.3.1

The Force on the Electron in Hydrogen

A hydrogen atom consists of a single proton and a single electron. The proton has a charge of $+e$ and the electron has $-e$ . In the “ground state” of the atom, the electron orbits the proton at most probable distance of $5.29\times10^{-11}~\mathrm{m}$ (Figure 1.3.2). Calculate the electric force on the electron due to the proton.

(Figure 1.3.2) $\begin{gather*}.\end{gather*}$

A positive charge is shown at the center of a sphere of radius r. An electron is depicted as a particle on the sphere. The force on the electron is along the radius, toward the nucleus.

Figure 1.3.2. A schematic depiction of a hydrogen atom, showing the force on the electron. This depiction is only to enable us to calculate the force; the hydrogen atom does not really look like this. Recall Figure 1.1.6.

Strategy

For the purposes of this example, we are treating the electron and proton as two point particles, each with an electric charge, and we are told the distance between them; we are asked to calculate the force on the electron. We thus use Coulomb’s law.

Solution

Our two charges and the distance between them are,

$q_1=+e=+1.602\times10^{-19}~\mathrm{C}$

$q_2=-e=-1.602\times10^{-19}~\mathrm{C}$

$r=5.29\times10^{-11}~m.$

The magnitude of the force on the electron is

$F=\frac{1}{4\pi\varepsilon_0}\frac{|e|^2}{r^2}=\frac{1}{4\pi\left(8.85\times10^{-12}~\frac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2}\right)}\frac{\left(1.602\times10^{-19}~\mathrm{C}\right)^2}{\left(5.29\times10^{-11}~m\right)^2}=8.25\times10^{-8}~\mathrm{N}.$

As for the direction, since the charges on the two particles are opposite, the force is attractive; the force on the electron points radially directly toward the proton, everywhere in the electron’s orbit. The force is thus expressed as

$\vec{\mathbf{F}}=(8.25\times10^{-8}~\mathrm{N})~\mathbf{\hat{r}}.$

Significance

This is a three-dimensional system, so the electron (and therefore the force on it) can be anywhere in an imaginary spherical shell around the proton. In this “classical” model of the hydrogen atom, the electrostatic force on the electron points in the inward centripetal direction, thus maintaining the electron’s orbit. But note that the quantum mechanical model of hydrogen is utterly different.

CHECK YOUR UNDERSTANDING 1.1

What would be different if the electron also had a positive charge?

Multiple Source Charges

The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat the analysis, two charges at a time. Specifically, we ask the question: Given $N$ charges (which we refer to as source charge), what is the net electric force that they exert on some other point charge (which we call the test charge)? Note that we use these terms because we can think of the test charge being used to test the strength of the force provided by the source charges.

Like all forces that we have seen up to now, the net electric force on our test charge is simply the vector sum of each individual electric force exerted on it by each of the individual test charges. Thus, we can calculate the net force on the test charge $Q$ by calculating the force on it from each source charge, taken one at a time, and then adding all those forces together (as vectors). This ability to simply add up individual forces in this way is referred to as the principle of superposition, and is one of the more important features of the electric force. In mathematical form, this becomes

(1.3.2) $\begin{equation*}{\mathbf{F}}=\frac{1}{4\pi\varepsilon_0}Q\sum_{i=1}^{N}\frac{q_i}{r_i^2}\mathbf{\hat{r}}_i.\end{equation*}$

In this expression, $Q$ represents the charge of the particle that is experiencing the electric force $\vec{\mathbf{F}}$ , and is located at $\vec{\mathbf{r}}$ from the origin; the $q_i$ are the $N$ source charges, and the vectors $\vec{\mathbf{r}}_i=r_i\hat{\mathbf{r}}_i$ are the displacements from the position of the $i$ th charge to the position of $Q$ . Each of the $N$ unit vectors points directly from its associated source charge toward the test charge. All of this is depicted in Figure 1.3.3. Please note that there is no physical difference between $Q$ and $q_i;$ the difference in labels is merely to allow clear discussion, with $Q$ being the charge we are determining the force on.

(Figure 1.3.3) $\begin{gather*}.\end{gather*}$

Eight source charges are shown as small spheres distributed within an x y z coordinate system. The sources are labeled q sub 1, q sub 2, and so on. Sources 1, 2, 4, 7 and 8 are shaded red and sources 3, 5, and 6 are shaded blue. A test charge is also shown, shaded in green and labeled as plus Q. The r vectors from each source to the test charge Q are shown as arrows with tails at the sources and heads at the test charge. The vector from q sub 1 to the test charge is labeled as r sub 1. The vector from q sub 2 to the test charge is labeled as r sub 2, and so on for all eight vectors.

Figure 1.3.3. The eight source charges each apply a force on the single test charge Q. Each force can be calculated independently of the other seven forces. This is the essence of the superposition principle.

(Note that the force vector $\vec{\mathbf{F}}_i$ does not necessarily point in the same direction as the unit vector $\hat{\mathbf{r}}_i$ ; it may point in the opposite direction, $-\hat{\mathbf{r}}_i$ . The signs of the source charge and test charge determine the direction of the force on the test charge.)

There is a complication, however. Just as the source charges each exert a force on the test charge, so too (by Newton’s third law) does the test charge exert an equal and opposite force on each of the source charges. As a consequence, each source charge would change position. However, by Equation 1.3.2, the force on the test charge is a function of position; thus, as the positions of the source charges change, the net force on the test charge necessarily changes, which changes the force, which again changes the positions. Thus, the entire mathematical analysis quickly becomes intractable. Later, we will learn techniques for handling this situation, but for now, we make the simplifying assumption that the source charges are fixed in place somehow, so that their positions are constant in time. (The test charge is allowed to move.) With this restriction in place, the analysis of charges is known as electrostatics, where “statics” refers to the constant (that is, static) positions of the source charges and the force is referred to as an electrostatic force.

EXAMPLE 1.3.2

The Net Force from Two Source Charges

Three different, small charged objects are placed as shown in Figure 1.3.4. The charges $q_1$ and $q_3$ are fixed in place; $q_2$ is free to move. Given $q_1=2e$ , $q_2=-3e$ , and $q_3=-5e$ , and that $d=2.0\times10^{-7}~\mathrm{m}$ what is the net force on the middle charge $q_2$ ?

(Figure 1.3.4) $\begin{gather*}.\end{gather*}$

Three charges are shown in an x y coordinate system. Charge q sub 1 is at x=0, y=d. Charge q sub 2 is at x=2 d, y=0. Charge q sub 3 is at the origin. Force F 1 2 is exerted on charge q sub 2 and points up. Force F 2 3 is exerted on charge q sub 2 and points to the left. Force F is exerted on charge q sub 2 and points at an angle theta above the minus x direction.

Figure 1.3.4. Source charges $q_1$ and $q_3$ each apply a force on $q_2$ .

Strategy

We use Coulomb’s law again. The way the question is phrased indicates that $q_2$ is our test charge, so that $q_1$ and $q_3$ are source charges. The principle of superposition says that the force on $q_2$ from each of the other charges is unaffected by the presence of the other charge. Therefore, we write down the force on $q_2$ from each and add them together as vectors.

Solution

We have two source charges ( $q_1$ and $q_3$ ), a test charge ( $q_2$ ), distances ( $r_{21}$ and $r_{23}$ ), and we are asked to find a force. This calls for Coulomb’s law and superposition of forces. There are two forces:

$\vec{\mathbf{F}}=\vec{\mathbf{F}}_{21}+\vec{\mathbf{F}}_{23}=\frac{1}{4\pi\epsilon_0}\left[\frac{q_2q_1}{r_{21}^2}\hat{\mathbf{j}}+\left(-\frac{q_2q_3}{r_{23}^2}\hat{\mathbf{i}}\right)\right].$

We can’t add these forces directly because they don’t point in the same direction: $\vec{\mathbf{F}}_{12}$ points only in the $-x$ -direction, while $\vec{\mathbf{F}}_{13}$ points only in the $+y$ -direction. The net force is obtained from applying the Pythagorean theorem to its $x-$ and $y$ -components:

$F=\sqrt{F_x^2+F_y^2}$

where

$\begin{eqnarray*} F_x&=&-F_{23}=-\frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_{23}^2}\\&=&-\left(8.99\times10^{9}~\mathrm{N}\cdot\frac{\mathrm{m}^2}{\mathrm{C}^2}\right)\frac{(4.806\times10^{-19}~\mathrm{C})(8.01\times10^{-19}~\mathrm{C})}{(4.00\times10^{-7}~\mathrm{m})^2}\\&=&-2.16\times10^{-14}~\mathrm{N}\end{eqnarray*}$

and

$\begin{eqnarray*} F_y&=&F_{21}=\frac{1}{4\pi\epsilon_0}\frac{q_2q_1}{r_{21}^2}\\&=&-\left(8.99\times10^{9}~\mathrm{N}\cdot\frac{\mathrm{m}^2}{\mathrm{C}^2}\right)\frac{(4.806\times10^{-19}~\mathrm{C})(3.024\times10^{-19}~\mathrm{C})}{(2.00\times10^{-7}~\mathrm{m})^2}\\&=&3.46\times10^{-14}~\mathrm{N}.\end{eqnarray*}$

We find that

$F=\sqrt{F_x^2+F_y^2}=4.08\times10^{-14}~\mathrm{N}$

at an angle of

$\phi=\tan^{-1}\left(\frac{F_y}{F_x}\right)=\tan^{-1}\left(\frac{3.46\times10^{-14}~\mathrm{N}}{-2.16\times10^{-14}~\mathrm{N}}\right)=-58^{\circ},$

that is, $58^{\circ}$ above the $-x$ -axis, as shown in the diagram.

Significance

Notice that when we substituted the numerical values of the charges, we did not include the negative sign of either $q_2$ or $q_3.$ Recall that negative signs on vector quantities indicate a reversal of direction of the vector in question. But for electric forces, the direction of the force is determined by the types (signs) of both interacting charges; we determine the force directions by considering whether the signs of the two charges are the same or are opposite. If you also include negative signs from negative charges when you substitute numbers, you run the risk of mathematically reversing the direction of the force you are calculating. Thus, the safest thing to do is to calculate just the magnitude of the force, using the absolute values of the charges, and determine the directions physically.

It’s also worth noting that the only new concept in this example is how to calculate the electric forces; everything else (getting the net force from its components, breaking the forces into their components, finding the direction of the net force) is the same as force problems you have done earlier.

CHECK YOUR UNDERSTANDING 1.2

What would be different if $q_1$ were negative?

Icon for the Creative Commons Attribution 4.0 International License

1.3 Coulomb's Law by Daryl Janzen is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

5

1.4 Electric Field

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Explain the purpose of the electric field concept
Describe the properties of the electric field
Calculate the field of a collection of source charges of either sign

As we showed in the preceding section, the net electric force on a test charge is the vector sum of all the electric forces acting on it, from all of the various source charges, located at their various positions. But what if we use a different test charge, one with a different magnitude, or sign, or both? Or suppose we have a dozen different test charges we wish to try at the same location? We would have to calculate the sum of the forces from scratch. Fortunately, it is possible to define a quantity, called the electric field, which is independent of the test charge. It only depends on the configuration of the source charges, and once found, allows us to calculate the force on any test charge.

Defining a Field

Suppose we have $N$ source charges $q_1,\ q_2,\ q_3,\ldots,\ q_N$ located at positions $\vec{\mathbf{r}}_1,\vec{\mathbf{r}}_2,\vec{\mathbf{r}}_3,\ldots,\vec{\mathbf{r}}_N$ , applying $N$ electrostatic forces on a test charge $Q$ . The net force on $Q$ is (see Equation 1.3.2)

$\begin{eqnarray*}\vec{\mathbf{F}}&=&\vec{\mathbf{F}}_1+\vec{\mathbf{F}}_2+\vec{\mathbf{F}}_3+\ldots+\vec{\mathbf{F}}_N\\&=&\frac{1}{4\pi\epsilon_0}\left(\frac{Qq_1}{r_1^2}\hat{\mathbf{r}}_1+\frac{Qq_2}{r_2^2}\hat{\mathbf{r}}_2+\frac{Qq_3}{r_3^2}\hat{\mathbf{r}}_3+\ldots+\frac{Qq_N}{r_N^2}\hat{\mathbf{r}}_N\right)\\&=&Q\left[\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_1^2}\hat{\mathbf{r}}_1+\frac{q_2}{r_2^2}\hat{\mathbf{r}}_2+\frac{q_3}{r_3^2}\hat{\mathbf{r}}_3+\ldots+\frac{q_N}{r_N^2}\hat{\mathbf{r}}_N\right)\right].\end{eqnarray*}$

We can rewrite this as

(1.4.1) $\begin{equation*}\vec{\mathbf{F}}=Q\vec{\mathbf{E}}\end{equation*}$

where

$\vec{\mathbf{E}}\equiv\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_1^2}\hat{\mathbf{r}}_1+\frac{q_2}{r_2^2}\hat{\mathbf{r}}_2+\frac{q_3}{r_3^2}\hat{\mathbf{r}}_3+\ldots+\frac{q_N}{r_N^2}\hat{\mathbf{r}}_N\right)$

or, more compactly,

\begin{equation*}\vec{\mathbf{E}}\equiv\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_i^2}\hat{\mathbf{r}}_i.\end{equation*}

This expression is called the electric field at position $P=P(x,y,z)$ of the $N$ source charges. Here, $P$ is the location of the point in space where you are calculating the field and is relative to the positions $\vec{\mathbf{r}}_i$ of the source charges (Figure 1.4.1). Note that we have to impose a coordinate system to solve actual problems.

(Figure 1.4.1) $\begin{gather*}.\end{gather*}$

Figure 1.4.1 Each of these eight source charges creates its own electric field at every point in space; shown here are the field vectors at an arbitrary point P. Like the electric force, the net electric field obeys the superposition principle.

Notice that the calculation of the electric field makes no reference to the test charge. Thus, the physically useful approach is to calculate the electric field and then use it to calculate the force on some test charge later, if needed. Different test charges experience different forces (Equation 1.4.1), but it is the same electric field (Equation 1.4.2). That being said, recall that there is no fundamental difference between a test charge and a source charge; these are merely convenient labels for the system of interest. Any charge produces an electric field; however, just as Earth’s orbit is not affected by Earth’s own gravity, a charge is not subject to a force due to the electric field it generates. Charges are only subject to forces from the electric fields of other charges.

In this respect, the electric field $\vec{\mathbf{E}}$ of a point charge is similar to the gravitational field $\vec{\mathbf{g}}$ of Earth; once we have calculated the gravitational field at some point in space, we can use it any time we want to calculate the resulting force on any mass we choose to place at that point. In fact, this is exactly what we do when we say the gravitational field of Earth (near Earth’s surface) has a value of $9.81~\mathrm{m/s}^2$ , and then we calculate the resulting force (i.e., weight) on different masses. Also, the general expression for calculating $\vec{\mathbf{g}}$ at arbitrary distances from the center of Earth (i.e., not just near Earth’s surface) is very similar to the expression for $\vec{\mathbf{E}}$ : $\vec{\mathbf{g}}=G\frac{M}{r^2}\hat{\mathbf{r}}$ , where $G$ is a proportionality constant, playing the same role for $\vec{\mathbf{g}}$ as $\frac{1}{4\pi\epsilon_0}$ does for $\vec{\mathbf{E}}$ . The value of $\vec{\mathbf{g}}$ is calculated once and is then used in an endless number of problems.

To push the analogy further, notice the units of the electric field: From $F=QE$ , the units of $E$ are newtons per coulomb, $\mathrm{N/C}$ , that is, the electric field applies a force on each unit charge. Now notice the units of $g$ : From $w=mg$ , the units of $g$ are newtons per kilogram, $\mathrm{N/kg}$ , that is, the gravitational field applies a force on each unit mass. We could say that the gravitational field of Earth, near Earth’s surface, has a value of $9.81~\mathrm{N/kg}$ .

The Meaning of “Field”

Recall from your studies of gravity that the word “field” in this context has a precise meaning. A field, in physics, is a physical quantity whose value depends on (is a function of) position, relative to the source of the field. In the case of the electric field, Equation 1.4.2 shows that the value of $\vec{\mathbf{E}}$ (both the magnitude and the direction) depends on where in space the point $P$ is located, measured from the locations $\vec{\mathbf{r}}_i$ of the source charges $q_i$ .

In addition, since the electric field is a vector quantity, the electric field is referred to as a vector field. (The gravitational field is also a vector field.) In contrast, a field that has only a magnitude at every point is a scalar field. The temperature in a room is an example of a scalar field. It is a field because the temperature, in general, is different at different locations in the room, and it is a scalar field because temperature is a scalar quantity.

Also, as you did with the gravitational field of an object with mass, you should picture the electric field of a charge-bearing object (the source charge) as a continuous, immaterial substance that surrounds the source charge, filling all of space—in principle, to $\pm\infty$ in all directions. The field exists at every physical point in space. To put it another way, the electric charge on an object alters the space around the charged object in such a way that all other electrically charged objects in space experience an electric force as a result of being in that field. The electric field, then, is the mechanism by which the electric properties of the source charge are transmitted to and through the rest of the universe. (Again, the range of the electric force is infinite.)

We will see in subsequent chapters that the speed at which electrical phenomena travel is the same as the speed of light. There is a deep connection between the electric field and light.

Superposition

Yet another experimental fact about the field is that it obeys the superposition principle. In this context, that means that we can (in principle) calculate the total electric field of many source charges by calculating the electric field of only $q_1$ at position $P$ , then calculate the field of $q_2$ at $P$ , while—and this is the crucial idea—ignoring the field of, and indeed even the existence of, $q_1$ . We can repeat this process, calculating the field of each individual source charge, independently of the existence of any of the other charges. The total electric field, then, is the vector sum of all these fields. That, in essence, is what Equation 1.4.2 says.

In the next section, we describe how to determine the shape of an electric field of a source charge distribution and how to sketch it.

The Direction of the Field

Equation 1.4.2 enables us to determine the magnitude of the electric field, but we need the direction also. We use the convention that the direction of any electric field vector is the same as the direction of the electric force vector that the field would apply to a positive test charge placed in that field. Such a charge would be repelled by positive source charges (the force on it would point away from the positive source charge) but attracted to negative charges (the force points toward the negative source).

DIRECTION OF THE ELECTRIC FIELD

By convention, all electric fields $\vec{\mathbf{E}}$ point away from positive source charges and point toward negative source charges.

INTERACTIVE

Add charges to the Electric Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

EXAMPLE 1.4.1

The E-field of an Atom

In an ionized helium atom, the most probable distance between the nucleus and the electron is $r=26.5\times10^{-12}~\mathrm{m}$ . What is the electric field due to the nucleus at the location of the electron?

Strategy

Note that although the electron is mentioned, it is not used in any calculation. The problem asks for an electric field, not a force; hence, there is only one charge involved, and the problem specifically asks for the field due to the nucleus. Thus, the electron is a red herring; only its distance matters. Also, since the distance between the two protons in the nucleus is much, much smaller than the distance of the electron from the nucleus, we can treat the two protons as a single charge $+2e$ (Figure 1.4.2).

(Figure 1.4.2) $\begin{gather*}.\end{gather*}$

A positive charge of plus 2 e is shown at the center of a sphere of radius r. An electron is depicted as a particle on the sphere. The vector r is shown as a vector with its tail at the center and its head at the location of the electron. The electric field at the location of the electron is shown as a vector E with its tail at the electron and pointing directly away from the center.

Figure 1.4.2 A schematic representation of a helium atom. Again, helium physically looks nothing like this, but this sort of diagram is helpful for calculating the electric field of the nucleus.

Solution

The electric field is calculated by

$\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_i^2}\hat{\mathbf{r}}_i.$

Since there is only one source charge (the nucleus), this expression simplifies to

$\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}.$

Here $q=2e=2\left(1.6\times10^{-19}~\mathrm{C}\right)$ (since there are two protons) and $r$ is given; substituting gives

$\vec{\mathbf{E}}=\frac{1}{4\pi\left(8.85\times10^{-12}~\frac{\mathrm{C}^2}{\mathrm{N}\cdot\mathrm{m}^2}\right)}\frac{2\left(1.6\times10^{-19}~\mathrm{C}\right)}{\left(26.5\times10^{-12}~\mathrm{m}\right)^2}\hat{\mathbf{r}}=4.1\times10^{12}~\frac{\mathrm{N}}{\mathrm{C}}\hat{\mathbf{r}}.$

The direction of $\vec{\mathbf{E}}$ is radially away from the nucleus in all directions. Why? Because a positive test charge placed in this field would accelerate radially away from the nucleus (since it is also positively charged), and again, the convention is that the direction of the electric field vector is defined in terms of the direction of the force it would apply to positive test charges.

EXAMPLE 1.4.2

The E-Field above Two Equal Charges

(a) Find the electric field (magnitude and direction) a distance $z$ above the midpoint between two equal charges $+q$ that are a distance $d$ apart (Figure 1.4.3). Check that your result is consistent with what you’d expect when $z\gg{d}$ .

(b) The same as part (a), only this time make the right-hand charge $-q$ instead of $+q$ .

(Figure 1.4.3) $\begin{gather*}.\end{gather*}$

Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta.

Figure 1.4.3 Finding the field of two identical source charges at the point P. Due to the symmetry, the net field at P is entirely vertical. (Notice that this is not true away from the midline between the charges.)

Strategy

We add the two fields as vectors, per Equation 1.4.2. Notice that the system (and therefore the field) is symmetrical about the vertical axis; as a result, the horizontal components of the field vectors cancel. This simplifies the math. Also, we take care to express our final answer in terms of only quantities that are given in the original statement of the problem: $q,\ z,\ d$ , and constants $(\pi,\ \epsilon_0)$ .

Solution

(a) By symmetry, the horizontal ( $x$ )-components of $\vec{\mathbf{E}}$ cancel (Figure 1.4.4);

$E_x=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sin\theta-\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sin\theta=0.$

(Figure 1.4.4) $\begin{gather*}.\end{gather*}$

Figure 1.4.4 Note that the horizontal components of the electric fields from the two charges cancel each other out, while the vertical components add together.

The vertical ( $z$ )-component is given by

$E_z=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta+\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta=\frac{1}{4\pi\epsilon_0}\frac{2q}{r^2}\cos\theta.$

Since none of the other components survive, this is the entire electric field, and it points in the $\hat{\mathbf{k}}$ -direction. Notice that this calculation uses the principle of superposition; we calculate the fields of the two charges independently and then add them together.

What we want to do now is replace the quantities in this expression that we don’t know (such as $r$ ), or can’t easily measure (such as $\cos\theta$ ) with quantities that we do know, or can measure. In this case, by geometry,

$r^2=z^2+\left(\frac{d}{2}\right)^2$

and

$\cos\theta=\frac{z}{r}=\frac{z}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{1/2}}.$

Thus, substituting,

$\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2q}{\left[z^2+\left(\frac{d}{2}\right)^2\right]}\frac{z}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{1/2}}\hat{\mathbf{k}}.$

Simplifying, the desired answer is

(1.4.3) $\begin{equation*}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2qz}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{3/2}}\hat{\mathbf{k}}.\end{equation*}$

(b) If the source charges are equal and opposite, the vertical components cancel because

$E_z=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta-\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\cos\theta=0.$

and we get, for the horizontal component of $\vec{\mathbf{E}}$ ,

$\begin{eqnarray*}\vec{\mathbf{E}}(z)&=&\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\sin\theta\hat{\mathbf{i}}-\frac{1}{4\pi\epsilon_0}\frac{-q}{r^2}\sin\theta\hat{\mathbf{i}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2q}{r^2}\sin\theta\hat{\mathbf{i}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2q}{\left[z^2+\left(\frac{d}{2}\right)^2\right]}\frac{\left(\frac{d}{2}\right)}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{1/2}}\hat{\mathbf{i}}.\end{eqnarray*}$

This becomes

(1.4.4) $\begin{equation*}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{qd}{\left[z^2+\left(\frac{d}{2}\right)^2\right]^{3/2}}\hat{\mathbf{i}}.\end{equation*}$

Significance

It is a very common and very useful technique in physics to check whether your answer is reasonable by evaluating it at extreme cases. In this example, we should evaluate the field expressions for the cases $d=0$ , $z\gg d$ , and $z\rightarrow\infty$ , and confirm that the resulting expressions match our physical expectations. Let’s do so:

Let’s start with Equation 1.4.3, the field of two identical charges. From far away (i.e., $z\gg d$ ), the two source charges should “merge” and we should then “see” the field of just one charge, of size $2q$ . So, let $z\gg d$ ; then we can neglect $d^2$ in Equation 1.4.3 to obtain

$\begin{eqnarray*}\lim_{d\rightarrow0}\vec{\mathbf{E}}&=&\frac{1}{4\pi\epsilon_0}\frac{2qz}{\left[z^2\right]^{3/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2qz}{z^3}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{2q}{z^2}\hat{\mathbf{k}},\end{eqnarray*}$

which is the correct expression for a field at a distance $z$ away from a charge $2q$ .

Next, we consider the field of equal and opposite charges, Equation 1.4.4. It can be shown (via a Taylor expansion) that for $d\ll z\ll\infty$ , this becomes

(1.4.5) $\begin{equation*}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{qd}{z^3}\hat{\mathbf{i}},\end{equation*}$

which is the field of a dipole, a system that we will study in more detail later. (Note that the units of $\vec{\mathbf{E}}$ are still correct in this expression, since the units of $d$ in the numerator cancel the unit of the “extra” $z$ in the denominator.) If $z$ is very large $(z\rightarrow\infty)$ , then $E\rightarrow0$ , as it should; the two charges “merge” and so cancel out.

CHECK YOUR UNDERSTANDING 1.3

What is the electric field due to a single point particle?

INTERACTIVE

Try this simulation of electric field hockey to get the charge in the goal by placing other charges on the field.

6

1.5 Calculating Electric Fields of Charge Distributions

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Explain what a continuous source charge distribution is and how it is related to the concept of quantization of charge
Describe line charges, surface charges, and volume charges
Calculate the field of a continuous source charge distribution of either sign

The charge distributions we have seen so far have been discrete: made up of individual point particles. This is in contrast with a continuous charge distribution, which has at least one nonzero dimension. If a charge distribution is continuous rather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge.

Note that because charge is quantized, there is no such thing as a “truly” continuous charge distribution. However, in most practical cases, the total charge creating the field involves such a huge number of discrete charges that we can safely ignore the discrete nature of the charge and consider it to be continuous. This is exactly the kind of approximation we make when we deal with a bucket of water as a continuous fluid, rather than a collection of H2O molecules.

Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, as shown in Figure 1.5.1.

(Figure 1.5.1) $\begin{gather*}.\end{gather*}$

Figure a shows a long rod with linear charge density lambda. A small segment of the rod is shaded and labeled d l. Figure b shows a surface with surface charge density sigma. A small area within the surface is shaded and labeled d A. Figure c shows a volume with volume charge density rho. A small volume within it is shaded and labeled d V. Figure d shows a surface with two regions shaded and labeled q 1 and q2. A point P is identified above (not on) the surface. A thin line indicates the distance from each of the shaded regions. The vectors E 1 and E 2 are drawn at point P and point away from the respective shaded region. E net is the vector sum of E 1 and E 2. In this case, it points up, away from the surface.

Figure 1.5.1 The configuration of charge differential elements for a (a) line charge, (b) sheet of charge, and (c) a volume of charge. Also note that (d) some of the components of the total electric field cancel out, with the remainder resulting in a net electric field.

Definitions of charge density:

$\lambda\equiv$ charge per unit length (linear charge density); units are coulombs per metre ( $\mathrm{C/m}$ )
$\sigma\equiv$ charge per unit area (surface charge density); units are coulombs per square metre ( $\mathrm{C/m^2}$ )
$\rho\equiv$ charge per unit volume (volume charge density); units are coulombs per cubic metre ( $\mathrm{C/m^3}$ )

Then, for a line charge, a surface charge, and a volume charge, the summation in Equation 1.4.2 becomes an integral and $q_i$ is replaced by $dq=\lambda dl$ , $\sigma dA$ , or $\rho dV$ respectively:

\begin{equation*}\mathrm{Point~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\left(\frac{q_i}{r^2}\right)\hat{\mathbf{r}}\end{equation*}

(1.5.2) $\begin{equation*}\mathrm{Line~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)\hat{\mathbf{r}}\end{equation*}$

(1.5.3) $\begin{equation*}\mathrm{Surface~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{surface}}\left(\frac{\sigma dA}{r^2}\right)\hat{\mathbf{r}}\end{equation*}$

(1.5.4) $\begin{equation*}\mathrm{Volume~charge:}&~~~~~&\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{volume}}\left(\frac{\rho dV}{r^2}\right)\hat{\mathbf{r}}\end{equation*}$

The integrals are generalizations of the expression for the field of a point charge. They implicitly include and assume the principle of superposition. The “trick” to using them is almost always in coming up with correct expressions for $dl$ , $dA$ , or $dV$ as the case may be, expressed in terms of $r$ , and also expressing the charge density function appropriately. It may be constant; it might be dependent on location.

Note carefully the meaning of $r$ in these equations: It is the distance from the charge element $(q_i,\lambda dl,\sigma dA,\rho dV)$ to the location of interest, $P(x,y,z)$ (the point in space where you want to determine the field). However, don’t confuse this with the meaning of $\hat{\mathbf{r}}$ ; we are using it and the vector notation $\vec{\mathbf{E}}$ to write three integrals at once. That is, Equation 1.5.2 is actually

$E_x(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)_x,$

$E_y(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)_y,$

$E_z(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\left(\frac{\lambda dl}{r^2}\right)_z.$

EXAMPLE 1.5.1

Electric Field of a Line Segment

Find the electric field a distance $z$ above the midpoint of a straight line segment of length $L$ that carries a uniform line charge density $\lambda$ .

Strategy

Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length $dl$ , each of which carries a differential amount of charge $dq=\lambda dl$ . Then, we calculate the differential field created by two symmetrically placed pieces of the wire, using the symmetry of the setup to simplify the calculation (Figure 1.5.2). Finally, we integrate this differential field expression over the length of the wire (half of it, actually, as we explain below) to obtain the complete electric field expression.

(Figure 1.5.2) $\begin{gather*}.\end{gather*}$

A long, thin wire is on the x axis. The end of the wire is a distance z from the center of the wire. A small segment of the wire, a distance x to the right of the center of the wire, is shaded. Another segment, the same distance to the left of center, is also shaded. Point P is a distance z above the center of the wire, on the z axis. Point P is a distance r from each shaded region. The r vectors point from each shaded region to point P. Vectors d E 1 and d E 2 are drawn at point P. d E 1 points away from the left side shaded region and points up and right, at an angle theta to the z axis. d E 2 points away from the right side shaded region and points up and r left, making the same angle with the vertical as d E 1. The two d E vectors are equal in length.

Figure 1.5.2 A uniformly charged segment of wire. The electric field at point $P$ can be found by applying the superposition principle to symmetrically placed charge elements and integrating.

Solution

Before we jump into it, what do we expect the field to “look like” from far away? Since it is a finite line segment, from far away, it should look like a point charge. We will check the expression we get to see if it meets this expectation.

The electric field for a line charge is given by the general expression

$\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}}\frac{\lambda dl}{r^2}\hat{\mathbf{r}}.$

The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal ( $x$ )-components of the field cancel, so that the net field points in the $z$ -direction. Let’s check this formally.

The total field $\vec{\mathbf{E}}(P)$ is the vector sum of the fields from each of the two charge elements (call them $\vec{\mathbf{E}}_1$ and $\vec{\mathbf{E}}_2$ , for now):

$\vec{\mathbf{E}}(P)=\vec{\mathbf{E}}_1+\vec{\mathbf{E}}_2=E_{ix}\hat{\mathbf{i}}+E_{1z}\hat{\mathbf{k}}+E_{2x}(-\hat{\mathbf{i}})+E_{2z}\hat{\mathbf{k}}.$

Because the two charge elements are identical and are the same distance away from the point $P$ where we want to calculate the field, $E_{1x}=E_{2x}$ , so those components cancel. This leaves

$\vec{\mathbf{E}}(P)=E_{1z}\hat{\mathbf{k}}+E_{2z}\hat{\mathbf{k}}=E_1\cos\theta\hat{\mathbf{k}}+E_2\cos\theta\hat{\mathbf{k}}.$

These components are also equal, so we have

$\begin{eqnarray*}\hat{\mathbf{E}}(P)&=&\frac{1}{4\pi\epsilon_0}\int\frac{\lambda dl}{r^2}\cos\theta\hat{\mathbf{k}}+\frac{1}{4\pi\epsilon_0}\int\frac{\lambda dl}{r^2}\cos\theta\hat{\mathbf{k}}\\&=&\int_0^{L/2}\frac{2\lambda dx}{r^2}\cos\theta\hat{\mathbf{k}}\end{eqnarray*}$

where our differential line element $dl$ is $dx$ , in this example, since we are integrating along a line of charge that lies on the $x$ -axis. (The limits of integration are $0$ to $\frac{L}{2}$ , not $-\frac{L}{2}$ to $+\frac{L}{2}$ , because we have constructed the net field from two differential pieces of charge $dq$ . If we integrated along the entire length, we would pick up an erroneous factor of $2$ .)

In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variables that are not given. In this case, both $r$ and $\theta$ change as we integrate outward to the end of the line charge, so those are the variables to get rid of. We can do that the same way we did for the two point charges: by noticing that

$r=\left(z^2+x^2\right)^{1/2}$

and

$\cos\theta=\frac{z}{r}=\frac{z}{\left(z^2+x^2\right)^{1/2}}.$

Substituting, we obtain

$\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\frac{1}{4\pi\epsilon_0}\int_0^{L/2}\frac{2\lambda dx}{\left(z^2+x^2\right)}\frac{z}{\left(z^2+x^2\right)^{1/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\int_0^{L/2}\frac{2\lambda z}{\left(z^2+x^2\right)^{3/2}}dx\hat{\mathbf{k}}\\&=&\frac{2\lambda z}{4\pi\epsilon_0}\left.\left[\frac{x}{z^2\sqrt{z^2+x^2}}\right]\right|_0^{L/2}\hat{\mathbf{k}}\end{eqnarray*}$

which simplifies to

(1.5.5) $\begin{equation*}\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\frac{\lambda L}{z\sqrt{z^2+\frac{L^2}{4}}}\hat{\mathbf{k}}.\end{equation*}$

Significance

Notice, once again, the use of symmetry to simplify the problem. This is a very common strategy for calculating electric fields. The fields of nonsymmetrical charge distributions have to be handled with multiple integrals and may need to be calculated numerically by a computer.

CHECK YOUR UNDERSTANDING 1.4

How would the strategy used above change to calculate the electric field at a point a distance $z$ above one end of the finite line segment?

EXAMPLE 1.5.2

Electric Field of an Infinite Line of Charge

Find the electric field a distance $z$ above the midpoint of an infinite line of charge that carries a uniform line charge density $\lambda$ .

Strategy

This is exactly like the preceding example, except the limits of integration will be $-\infty$ to $+\infty$ .

Solution

Again, the horizontal components cancel out, so we wind up with

$\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{\lambda dx}{r^2}\cos\theta\hat{\mathbf{k}}$

where our differential line element $dl$ is $dx$ , in this example, since we are integrating along a line of charge that lies on the $x$ -axis. Again,

$\cos\theta=\frac{z}{r}=\frac{z}{\left(z^2+x^2\right)^{1/2}}.$

Substituting, we obtain

$\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\frac{1}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{\lambda dx}{\left(z^2+x^2\right)}\frac{z}{\left(z^2+x^2\right)^{1/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}\int_{-\infty}^{\infty}\frac{\lambda z}{\left(z^2+x^2\right)^{3/2}}dx\hat{\mathbf{k}}\\&=&\frac{\lambda z}{4\pi\epsilon_0}\left.\left[\frac{x}{z^2\sqrt{z^2+x^2}}\right]\right|_{-\infty}^{\infty}\hat{\mathbf{k}},\end{eqnarray*}$

which simplifies to

$\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\frac{2\lambda}{z}\hat{\mathbf{k}}.$

Significance

Our strategy for working with continuous charge distributions also gives useful results for charges with infinite dimension.

In the case of a finite line of charge, note that for $z\gg L$ , $z^2$ dominates the $L$ in the denominator, so that Equation 1.5.5 simplifies to

$\vec{\mathbf{E}}\approx\frac{1}{4\pi\epsilon_0}\frac{\lambda L}{z^2}\hat{\mathbf{k}}.$

If you recall that $\lambda L=q$ , the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.

In the limit $L\rightarrow\infty$ , on the other hand, we get the field of an infinite straight wire, which is a straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated:

\begin{equation*}\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2\lambda}{z}\hat{\mathbf{k}}.\end{equation*}

An interesting artifact of this infinite limit is that we have lost the usual $1/r^2$ dependence that we are used to. This will become even more intriguing in the case of an infinite plane.

EXAMPLE 1.5.3

Electric Field due to a Ring of Charge

A ring has a uniform charge density $\lambda$ , with units of coulomb per unit meter of arc. Find the electric potential at a point on the axis passing through the center of the ring.

Strategy

We use the same procedure as for the charged wire. The difference here is that the charge is distributed on a circle. We divide the circle into infinitesimal elements shaped as arcs on the circle and use polar coordinates shown in Figure 1.5.3.

(Figure 1.5.3) $\begin{gather*}.\end{gather*}$

A ring of radius R is shown in the x y plane of an x y z coordinate system. The ring is centered on the origin. A small segment of the ring is shaded. The segment is at an angle of theta from the x axis, subtends an angle of d theta, and contains a charge of d q equal to lambda R d theta. Point P is on the z axis, a distance of z above the center of the ring. The distance from the shaded segment to point P is equal to the square root of R squared plus squared.

Figure 1.5.3 The system and variable for calculating the electric field due to a ring of charge.

Solution

The electric field for a line charge is given by the general expression

$\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{line}\frac{\lambda dl}{r^2}\hat{\mathbf{r}}.$

A general element of the arc between $\theta$ and $\theta+d\theta$ is of length $Rd\theta$ and therefore contains a charge equal to $\lambda Rd\theta$ . The element is at a distance of $r=\sqrt{z^2+R^2}$ from $P$ , the angle is $\cos\phi=\frac{z}{\sqrt{z^2+R^2}}$ , and therefore the electric field is

$\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\int_{\mathrm{line}}\frac{\lambda dl}{r^2}\hat{\mathbf{r}}=\frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\frac{\lambda Rd\theta}{z^2+R^2}\frac{z}{\sqrt{z^2+R^2}}\hat{\mathbf{z}}\\&=&\frac{1}{4\pi\epsilon_0}\frac{\lambda Rz}{\left(z^2+R^2\right)^{3/2}}\hat{\mathbf{z}}\int_0^{2\pi}d\theta=\frac{1}{4\pi\epsilon_0}\frac{2\pi\lambda Rz}{\left(z^2+R^2\right)^{3/2}}\hat{\mathbf{z}}\\&=&=\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{tot}}z}{\left(z^2+R^2\right)^{3/2}}\hat{\mathbf{z}}.\end{eqnarray*}$

Significance

As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when we take the limit of $z\gg R$ , we find that

$\vec{\mathbf{E}}\approx\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{tot}}}{z^2}\hat{\mathbf{z}},$

as we expect.

EXAMPLE 1.5.4

The Field of a Disk

Find the electric field of a circular thin disk of radius $R$ and uniform charge density at a distance $z$ above the centre of the disk (Figure 1.5.4).

(Figure 1.5.4) $\begin{gather*}.\end{gather*}$

A disk of radius R is shown in the x y plane of an x y z coordinate system. The disk is centered on the origin. A ring, concentric with the disk, of radius r prime and width d r prime is indicated and two small segments on opposite sides of the ring are shaded and labeled as having charge d q. The test point is on the z axis, a distance of z above the center of the disk. The distance from each shaded segment to the test point is r. The electric field contributions, d E, due to the d q charges are shown as arrows in the directions of the associated r vectors. The d E vectors are at an angle of theta to the z axis.

Figure 1.5.4 A uniformly charged disk. As in the line charge example, the field above the center of this disk can be calculated by taking advantage of the symmetry of the charge distribution.

Strategy

The electric field for a surface charge is given by

$\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{surface}}\frac{\sigma dA}{r^2}\hat{\mathbf{r}}.$

To solve surface charge problems, we break the surface into symmetrical differential “stripes” that match the shape of the surface; here, we’ll use rings, as shown in the figure. Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical $\hat{\mathbf{k}}$ -direction. The vertical component of the electric field is extracted by multiplying by $\cos\theta$ , so

$\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\int_{\mathrm{surface}}\frac{\sigma dA}{r^2}\cos\theta\hat{\mathbf{k}}.$

As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities. In this case,

$\begin{eqnarray*}dA&=&2\pi r'dr'\\r^2&=&r'^2+z^2\\\cos\theta&=&\frac{z}{\left(r'^2+z^2\right)^{1/2}}.\end{eqnarray*}$

(Please take note of the two different “ $r$ s” here; $r$ is the distance from the differential ring of charge to the point $P$ where we wish to determine the field, whereas $r'$ is the distance from the centre of the disk to the differential ring of charge.) Also, we already performed the polar angle integral in writing down $dA$ .

Solution

Substituting all this in, we get

$\begin{eqnarray*}\vec{\mathbf{E}}(P)&=&\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\int_0^R\frac{\sigma(2\pi r'dr')z}{(r'^2+z^2)^{3/2}}\hat{\mathbf{k}}\\&=&\frac{1}{4\pi\epsilon_0}(2\pi\sigma z)\left(\frac{1}{z}-\frac{1}{\sqrt{R^2+z^2}}\right)\hat{\mathbf{k}}\end{eqnarray*}$

or, more simply,

(1.5.7) $\begin{equation*}\hat{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\left(2\pi\sigma-\frac{2\pi\sigma z}{\sqrt{R^2+z^2}}\right)\hat{\mathbf{k}}.\end{equation*}$

Significance

Again, it can be shown (via a Taylor expansion) that when $z\gg R$ , this reduces to

$\vec{\mathbf{E}}(z)\approx\frac{1}{4\pi\epsilon_0}\frac{\sigma\pi R^2}{z^2}\hat{\mathbf{k}},$

which is the expression for a point charge $Q=\sigma\pi R^2$ .

CHECK YOUR UNDERSTANDING 1.5

How would the above limit change with a uniformly charged rectangle instead of a disk?

As $R\rightarrow\infty$ , Equation 1.5.7 reduces to the field of an infinite plane, which is a flat sheet whose area is much, much greater than its thickness, and also much, much greater than the distance at which the field is to be calculated:

\begin{equation*}\vec{\mathbf{E}}=\frac{\sigma}{2\epsilon_0}\hat{\mathbf{k}}\end{equation*}

Note that this field is constant. This surprising result is, again, an artifact of our limit, although one that we will make use of repeatedly in the future. To understand why this happens, imagine being placed above an infinite plane of constant charge. Does the plane look any different if you vary your altitude? No—you still see the plane going off to infinity, no matter how far you are from it. It is important to note that Equation 1.5.8 is because we are above the plane. If we were below, the field would point in the $-\hat{\mathbf{k}}$ -direction.

EXAMPLE 1.5.5

The Field of Two Infinite Planes

Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities (Figure 1.5.5).

(Figure 1.5.5) $\begin{gather*}.\end{gather*}$

The figure shows two vertically oriented parallel plates A and B separated by a distance d. Plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved outward at the ends of the plates. A charge q is moved from A to B. The work done W equals q times V sub A B, and the electric field intensity E equals V sub A B over d.

Figure 1.5.5 Two charged infinite planes. Note the direction of the electric field.

Strategy

We already know the electric field resulting from a single infinite plane, so we may use the principle of superposition to find the field from two.

Solution

The electric field points away from the positively charged plane and toward the negatively charged plane. Since the $\sigma$ are equal and opposite, this means that in the region outside of the two planes, the electric fields cancel each other out to zero.

However, in the region between the planes, the electric fields add, and we get

$\vec{\mathbf{E}}=\frac{\sigma}{\epsilon_0}\hat{\mathbf{i}}$

for the electric field. The $\hat{\mathbf{i}}$ is because in the figure, the field is pointing in the $+x$ -direction.

Significance

Systems that may be approximated as two infinite planes of this sort provide a useful means of creating uniform electric fields.

CHECK YOUR UNDERSTANDING 1.6

What would the electric field look like in a system with two parallel positively charged planes with equal charge densities?

7

1.6 Electric Field Lines

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Explain the purpose of an electric field diagram
Describe the relationship between a vector diagram and a field line diagram
Explain the rules for creating a field diagram and why these rules make physical sense
Sketch the field of an arbitrary source charge

Now that we have some experience calculating electric fields, let’s try to gain some insight into the geometry of electric fields. As mentioned earlier, our model is that the charge on an object (the source charge) alters space in the region around it in such a way that when another charged object (the test charge) is placed in that region of space, that test charge experiences an electric force. The concept of electric field lines, and of electric field line diagrams, enables us to visualize the way in which the space is altered, allowing us to visualize the field. The purpose of this section is to enable you to create sketches of this geometry, so we will list the specific steps and rules involved in creating an accurate and useful sketch of an electric field.

It is important to remember that electric fields are three-dimensional. Although in this book we include some pseudo-three-dimensional images, several of the diagrams that you’ll see (both here, and in subsequent chapters) will be two-dimensional projections, or cross-sections. Always keep in mind that in fact, you’re looking at a three-dimensional phenomenon.

Our starting point is the physical fact that the electric field of the source charge causes a test charge in that field to experience a force. By definition, electric field vectors point in the same direction as the electric force that a (hypothetical) positive test charge would experience, if placed in the field (Figure 1.6.1 ).

(Figure 1.6.1) $\begin{gather*}.\end{gather*}$

The electric field is shown as arrows at test points on a grid. In figure a, the field is shown in the x y plane, with x and y measured in meters and ranging from -4 meters to 4 meters. The arrows point away from the origin, and are largest near the origin, decreasing with distance from the origin. In figure b, a three dimensional vector field is shown. The charge is at the center and, again, the arrows are largest near the origin, decreasing with distance from the origin.

Figure 1.6.1 The electric field of a positive point charge. A large number of field vectors are shown. Like all vector arrows, the length of each vector is proportional to the magnitude of the field at each point. (a) Field in two dimensions; (b) field in three dimensions.

We’ve plotted many field vectors in the figure, which are distributed uniformly around the source charge. Since the electric field is a vector, the arrows that we draw correspond at every point in space to both the magnitude and the direction of the field at that point. As always, the length of the arrow that we draw corresponds to the magnitude of the field vector at that point. For a point source charge, the length decreases by the square of the distance from the source charge. In addition, the direction of the field vector is radially away from the source charge, because the direction of the electric field is defined by the direction of the force that a positive test charge would experience in that field. (Again, keep in mind that the actual field is three-dimensional; there are also field lines pointing out of and into the page.)

This diagram is correct, but it becomes less useful as the source charge distribution becomes more complicated. For example, consider the vector field diagram of a dipole (Figure 1.6.2 ).

(Figure 1.6.2) $\begin{gather*}.\end{gather*}$

A vector plot of the electric field due to two sources. The sources are not shown. The field is represented by arrows in an x y graph. Both x and y are in meters and both scales are from -2 meters to 4 meters. Near the origin, the arrows are long and point away from it. Near the point at coordinates 2, 0 the arrows are long and point toward the point. The arrows get smaller as we move farther from those two location and point in intermediate directions.

Figure 1.6.2 The vector field of a dipole. Even with just two identical charges, the vector field diagram becomes difficult to understand.

There is a more useful way to present the same information. Rather than drawing a large number of increasingly smaller vector arrows, we instead connect all of them together, forming continuous lines and curves, as shown in Figure 1.6.3 .

(Figure 1.6.3) $\begin{gather*}.\end{gather*}$

In part a, electric field lines emanating from a positive charge are shown as straight arrows radiating out from the charge in all directions. In part b, a pair of charges is shown, with one positive and the other negative. The field lines are represented by curved arrows. The arrows start from the positive charge, radiating outward but curving to end at the negative charge. The outer field lines extend beyond the drawing region, but follow the same behavior as those that are within the drawing area.

Figure 1.6.3 (a) The electric field line diagram of a positive point charge. (b) The field line diagram of a dipole. In both diagrams, the magnitude of the field is indicated by the field line density. The field vectors (not shown here) are everywhere tangent to the field lines.

Although it may not be obvious at first glance, these field diagrams convey the same information about the electric field as do the vector diagrams. First, the direction of the field at every point is simply the direction of the field vector at that same point. In other words, at any point in space, the field vector at each point is tangent to the field line at that same point. The arrowhead placed on a field line indicates its direction.

As for the magnitude of the field, that is indicated by the field line density—that is, the number of field lines per unit area passing through a small cross-sectional area perpendicular to the electric field. This field line density is drawn to be proportional to the magnitude of the field at that cross-section. As a result, if the field lines are close together (that is, the field line density is greater), this indicates that the magnitude of the field is large at that point. If the field lines are far apart at the cross-section, this indicates the magnitude of the field is small. Figure 1.6.4 shows the idea.

(Figure 1.6.4) $\begin{gather*}.\end{gather*}$

Seven electric field lines are shown, generally going from bottom left to top right. The field lines get closer together toward the top. Two square areas, perpendicular to the field lines, are shaded. All of the field lines pass through each shaded area. The area toward the top is smaller than the area toward the bottom.

Figure 1.6.4 Electric field lines passing through imaginary areas. Since the number of lines passing through each area is the same, but the areas themselves are different, the field line density is different. This indicates different magnitudes of the electric field at these points.

In Figure 1.6.4, the same number of field lines passes through both surfaces ( $S$ and $S'$ ), but the surface $S$ is larger than surface $S'$ . Therefore, the density of field lines (number of lines per unit area) is larger at the location of $S'$ , indicating that the electric field is stronger at the location of $S'$ than at $S$ . The rules for creating an electric field diagram are as follows.

Problem-Solving Strategy: Drawing Electric Field Lines

Electric field lines either originate on positive charges or come in from infinity, and either terminate on negative charges or extend out to infinity.
The number of field lines originating or terminating at a charge is proportional to the magnitude of that charge. A charge of $2q$ will have twice as many lines as a charge of $q$ .
At every point in space, the field vector at that point is tangent to the field line at that same point.
The field line density at any point in space is proportional to (and therefore is representative of) the magnitude of the field at that point in space.
Field lines can never cross. Since a field line represents the direction of the field at a given point, if two field lines crossed at some point, that would imply that the electric field was pointing in two different directions at a single point. This in turn would suggest that the (net) force on a test charge placed at that point would point in two different directions. Since this is obviously impossible, it follows that field lines must never cross.

Always keep in mind that field lines serve only as a convenient way to visualize the electric field; they are not physical entities. Although the direction and relative intensity of the electric field can be deduced from a set of field lines, the lines can also be misleading. For example, the field lines drawn to represent the electric field in a region must, by necessity, be discrete. However, the actual electric field in that region exists at every point in space.

Field lines for three groups of discrete charges are shown in Figure 1.6.5. Since the charges in parts (a) and (b) have the same magnitude, the same number of field lines are shown starting from or terminating on each charge. In (c), however, we draw three times as many field lines leaving the $+3q$ charge as entering the $-q$ . The field lines that do not terminate at $-q$ emanate outward from the charge configuration, to infinity.

(Figure 1.6.5) $\begin{gather*}.\end{gather*}$

Three pairs of charges and their field lines are shown. The charge on the left is positive in each case. In part a, the charge on the right is negative. The field lines are represented by curved arrows starting at the positive charge on the left, curving toward and terminating at the negative charge on the right. Between the charges, the field lines are dense. In part b, the charge on the right is positive. The field lines represented by curved arrows start at each of the positive charges and diverge outward. Between the charges, the field lines are less dense, and there is a black region midway between the charges. In part c, the charge on the right is negative. The field lines start at the positive charge. Some of the lines, those that start closest to the negative charge, curve toward the negative charge and terminate there. Lines that start further from the negative charge curve toward it but then diverge outward. There is an area with very low density of lines to the right of the pair of charges.

Figure 1.6.5 Three typical electric field diagrams. (a) A dipole. (b) Two identical charges. (c) Two charges with opposite signs and different magnitudes. Can you tell from the diagram which charge has the larger magnitude?

The ability to construct an accurate electric field diagram is an important, useful skill; it makes it much easier to estimate, predict, and therefore calculate the electric field of a source charge. The best way to develop this skill is with software that allows you to place source charges and then will draw the net field upon request. We strongly urge you to search the Internet for a program. Once you’ve found one you like, run several simulations to get the essential ideas of field diagram construction. Then practice drawing field diagrams, and checking your predictions with the computer-drawn diagrams.

INTERACTIVE

One example of a field-line drawing program is from the PhET “Charges and Fields” simulation.

8

1.7 Electric Dipoles

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Describe a permanent dipole
Describe an induced dipole
Define and calculate an electric dipole moment
Explain the physical meaning of the dipole moment

Earlier we discussed, and calculated, the electric field of a dipole: two equal and opposite charges that are “close” to each other. (In this context, “close” means that the distance $d$ between the two charges is much, much less than the distance of the field point $P$ , the location where you are calculating the field.) Let’s now consider what happens to a dipole when it is placed in an external field $\vec{\mathbf{E}}$ . We assume that the dipole is a permanent dipole; it exists without the field, and does not break apart in the external field.

Rotation of a Dipole due to an Electric Field

For now, we deal with only the simplest case: The external field is uniform in space. Suppose we have the situation depicted in Figure 1.7.1 , where we denote the distance between the charges as the vector $\vec{\mathbf{d}}$ , pointing from the negative charge to the positive charge. The forces on the two charges are equal and opposite, so there is no net force on the dipole. However, there is a torque:

$\begin{eqnarray*}\vec{\pmb{\uptau}}&=&\left(\frac{\vec{\mathbf{d}}}{2}\times\vec{\mathbf{F}}_+\right)+\left(-\frac{\vec{\mathbf{d}}}{2}\times\vec{\mathbf{F}}_-\right)\\&=&\left[\left(\frac{\vec{\mathbf{d}}}{2}\right)\times\left(+q\vec{\mathbf{E}}\right)+\left(-\frac{\vec{\mathbf{d}}}{2}\right)\times\left(-q\vec{\mathbf{E}}\right)\right]\\&=&q\vec{\mathbf{d}}\times\vec{\mathbf{E}}.\end{eqnarray*}$

(Figure 1.7.1) $\begin{gather*}.\end{gather*}$

In figure a dipole in a uniform electric field is shown along with the forces on the charges that make up the dipole. The dipole consists of a charge, minus q, and a positive charge, plus q, separated by a distance d. The line connecting the charges is at an angle to the horizontal so that the negative charge is above and to the left of the positive charge. The electric field E is horizontal and points to the right. The force on the negative charge is to the left, and is labeled as F minus. The force on the positive charge is to the right, and is labeled as F plus. Figure b shows the same diagram with the addition of the dipole moment vector, p, which points along the line connecting the charges, from the negative to the positive charge.

Figure 1.7.1 A dipole in an external electric field. (a) The net force on the dipole is zero, but the net torque is not. As a result, the dipole rotates, becoming aligned with the external field. (b) The dipole moment is a convenient way to characterize this effect. The $\vec{\mathbf{d}}$ points in the same direction as $\vec{\mathbf{p}}$ .

The quantity $q\vec{\mathbf{d}}$ (the magnitude of each charge multiplied by the vector distance between them) is a property of the dipole; its value, as you can see, determines the torque that the dipole experiences in the external field. It is useful, therefore, to define this product as the so-called dipole moment of the dipole:

\begin{equation*}\vec{\mathbf{p}}\equiv q\vec{\mathbf{d}}.\end{equation*}

We can therefore write

\begin{equation*}\vec{\pmb{\uptau}}=\vec{\mathbf{p}}\times\vec{\mathbf{E}}.\end{equation*}

Recall that a torque changes the angular velocity of an object, the dipole, in this case. In this situation, the effect is to rotate the dipole (that is, align the direction of $\vec{\mathbf{p}}$ ) so that it is parallel to the direction of the external field.

Induced Dipoles

Neutral atoms are, by definition, electrically neutral; they have equal amounts of positive and negative charge. Furthermore, since they are spherically symmetrical, they do not have a “built-in” dipole moment the way most asymmetrical molecules do. They obtain one, however, when placed in an external electric field, because the external field causes oppositely directed forces on the positive nucleus of the atom versus the negative electrons that surround the nucleus. The result is a new charge distribution of the atom, and therefore, an induced dipole moment (Figure 1.7.2).

(Figure 1.7.2) $\begin{gather*}.\end{gather*}$

Figure a illustrates a simplified model of a neutral atom. The nucleus is at the center of a uniform sphere negative charge. Figure b shows the atom in a horizontal, uniform electric field, E, that points to the right. The nucleus has shifted to the right a distance d, so that it is no longer at the center of the electron sphere. The result is an induced dipole moment, p, pointing to the right.

Figure 1.7.2 A dipole is induced in a neutral atom by an external electric field. The induced dipole moment is aligned with the external field.

An important fact here is that, just as for a rotated polar molecule, the result is that the dipole moment ends up aligned parallel to the external electric field. Generally, the magnitude of an induced dipole is much smaller than that of an inherent dipole. For both kinds of dipoles, notice that once the alignment of the dipole (rotated or induced) is complete, the net effect is to decrease the total electric field $\vec{\mathbf{E}}_{\mathrm{total}}=\vec{\mathbf{E}}_{\mathrm{external}}+\vec{\mathbf{E}}_{\mathrm{dipole}}$ in the regions outside the dipole charges (Figure 1.7.3 ). By “outside” we mean further from the charges than they are from each other. This effect is crucial for capacitors, as you will see in Capacitance.

(Figure 1.7.3) $\begin{gather*}.\end{gather*}$

A dipole, consisting of a negative charge on the left and a positive charge on the right is in a uniform electric field pointing to the right. The dipole moment, p, points to the right. The field lines of the net electric field are the sum of the dipole field and the uniform external field, horizontal far from the dipole and similar to the dipole field near the dipole.

Figure 1.7.3 The net electric field is the vector sum of the field of the dipole plus the external field.

Recall that we found the electric field of a dipole in Equation 1.4.5. If we rewrite it in terms of the dipole moment we get:

$\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{\vec{\mathbf{p}}}{z^3}.$

The form of this field is shown in Figure 1.7.3. Notice that along the plane perpendicular to the axis of the dipole and midway between the charges, the direction of the electric field is opposite that of the dipole and gets weaker the further from the axis one goes. Similarly, on the axis of the dipole (but outside it), the field points in the same direction as the dipole, again getting weaker the further one gets from the charges.

9

Chapter 1 Review

Key Terms

charging by induction
process by which an electrically charged object brought near a neutral object creates a charge separation in that object

conduction electron
electron that is free to move away from its atomic orbit

conductor
material that allows electrons to move separately from their atomic orbits; object with properties that allow charges to move about freely within it

continuous charge distribution
total source charge composed of so large a number of elementary charges that it must be treated as continuous, rather than discrete

coulomb
SI unit of electric charge

Coulomb force
another term for the electrostatic force

Coulomb’s law
mathematical equation calculating the electrostatic force vector between two charged particles

dipole
two equal and opposite charges that are fixed close to each other

dipole moment
property of a dipole; it characterizes the combination of distance between the opposite charges, and the magnitude of the charges

electric charge
physical property of an object that causes it to be attracted toward or repelled from another charged object; each charged object generates and is influenced by a force called an electric force

electric field
physical phenomenon created by a charge; it “transmits” a force between a two charges

electric force
noncontact force observed between electrically charged objects

electron
particle surrounding the nucleus of an atom and carrying the smallest unit of negative charge

electrostatic attraction
phenomenon of two objects with opposite charges attracting each other

electrostatic force
amount and direction of attraction or repulsion between two charged bodies; the assumption is that the source charges remain motionless

electrostatic repulsion
phenomenon of two objects with like charges repelling each other

electrostatics
study of charged objects which are not in motion

field line
smooth, usually curved line that indicates the direction of the electric field

field line density
number of field lines per square meter passing through an imaginary area; its purpose is to indicate the field strength at different points in space

induced dipole
typically an atom, or a spherically symmetric molecule; a dipole created due to opposite forces displacing the positive and negative charges

infinite plane
flat sheet in which the dimensions making up the area are much, much greater than its thickness, and also much, much greater than the distance at which the field is to be calculated; its field is constant

infinite straight wire
straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated

insulator
material that holds electrons securely within their atomic orbits

ion
atom or molecule with more or fewer electrons than protons

law of conservation of charge
net electric charge of a closed system is constant

linear charge density
amount of charge in an element of a charge distribution that is essentially one-dimensional (the width and height are much, much smaller than its length); its units are $\mathrm{C/m}$

neutron
neutral particle in the nucleus of an atom, with (nearly) the same mass as a proton

permanent dipole
typically a molecule; a dipole created by the arrangement of the charged particles from which the dipole is created

permittivity of vacuum
also called the permittivity of free space, and constant describing the strength of the electric force in a vacuum

polarization
slight shifting of positive and negative charges to opposite sides of an object

principle of superposition
useful fact that we can simply add up all of the forces due to charges acting on an object

proton
particle in the nucleus of an atom and carrying a positive charge equal in magnitude to the amount of negative charge carried by an electron

static electricity
buildup of electric charge on the surface of an object; the arrangement of the charge remains constant (“static”)

superposition
concept that states that the net electric field of multiple source charges is the vector sum of the field of each source charge calculated individually

surface charge density
amount of charge in an element of a two-dimensional charge distribution (the thickness is small); its units are $\mathrm{C/m}^2$

volume charge density
amount of charge in an element of a three-dimensional charge distribution; its units are $\mathrm{C/m}^3$

Key Equations

Coulomb’s law	$\vec{\mathbf{F}}_{12}(r)=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}\hat{\mathbf{r}}_{12}$
Superposition of electric forces	$\vec{\mathbf{F}}(r)=\frac{1}{4\pi\epsilon_0}Q\sum_{i=1}^{N}\frac{q_i}{r_{i}^2}\hat{\mathbf{r}}_{i}$
Electric force due to an electric field	$\vec{\mathbf{F}}=Q\vec{\mathbf{E}}$
Electric field at point $P$	$\vec{\mathbf{E}}(P)=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_{i}^2}\hat{\mathbf{r}}_{i}$
Field of an infinite wire	$\vec{\mathbf{E}}(z)=\frac{1}{4\pi\epsilon_0}\frac{2\lambda}{z}\hat{\mathbf{k}}$
Field of an infinite plane	$\vec{\mathbf{E}}=\frac{\sigma}{2\epsilon_0}\hat{\mathbf{k}}$
Dipole moment	$\vec{\mathbf{p}}=q\vec{\mathbf{d}}$
Torque on dipole in external $E$ -field	$\vec{\mathbf{\tau}}=\vec{\mathbf{p}}\times\vec{\mathbf{E}}$

Summary

1.1 Electric Charge

There are only two types of charge, which we call positive and negative. Like charges repel, unlike charges attract, and the force between charges decreases with the square of the distance.
The vast majority of positive charge in nature is carried by protons, whereas the vast majority of negative charge is carried by electrons. The electric charge of one electron is equal in magnitude and opposite in sign to the charge of one proton.
An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons.
The SI unit for charge is the coulomb ( $\mathrm{C}$ ), with protons and electrons having charges of opposite sign but equal magnitude; the magnitude of this basic charge is $e=1.602\times10^{-19}~\mathrm{C}$
Both positive and negative charges exist in neutral objects and can be separated by bringing the two objects into physical contact; rubbing the objects together can remove electrons from the bonds in one object and place them on the other object, increasing the charge separation.
For macroscopic objects, negatively charged means an excess of electrons and positively charged means a depletion of electrons.
The law of conservation of charge states that the net charge of a closed system is constant.

1.2 Conductors, Insulators, and Charging by Induction

A conductor is a substance that allows charge to flow freely through its atomic structure.
An insulator holds charge fixed in place.
Polarization is the separation of positive and negative charges in a neutral object. Polarized objects have their positive and negative charges concentrated in different areas, giving them a charge distribution.

1.3 Coulomb’s Law

Coulomb’s law gives the magnitude of the force between point charges. It is
$\vec{\mathbf{F}}_{12}(r)=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}^2}\hat{\mathbf{r}}_{12}$

where $q_1$ and $q_2$ are two point charges separated by a distance $r$ . This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces.

1.4 Electric Field

The electric field is an alteration of space caused by the presence of an electric charge. The electric field mediates the electric force between a source charge and a test charge.
The electric field, like the electric force, obeys the superposition principle
The field is a vector; by definition, it points away from positive charges and toward negative charges.

1.5 Calculating Electric Fields of Charge Distributions

A very large number of charges can be treated as a continuous charge distribution, where the calculation of the field requires integration. Common cases are:
- one-dimensional (like a wire); uses a line charge density $\lambda$
- two-dimensional (metal plate); uses surface charge density $\sigma$
- three-dimensional (metal sphere); uses volume charge density $\rho$
The “source charge” is a differential amount of charge $dq$ . Calculating $dq$ depends on the type of source charge distribution:
$dq=\lambda dl;~~dq=\sigma dA;~~dq=\rho dV.$
Symmetry of the charge distribution is usually key.
Important special cases are the field of an “infinite” wire and the field of an “infinite” plane.

1.6 Electric Field Lines

Electric field diagrams assist in visualizing the field of a source charge.
The magnitude of the field is proportional to the field line density.
Field vectors are everywhere tangent to field lines.

1.7 Electric Dipoles

If a permanent dipole is placed in an external electric field, it results in a torque that aligns it with the external field.
If a nonpolar atom (or molecule) is placed in an external field, it gains an induced dipole that is aligned with the external field.
The net field is the vector sum of the external field plus the field of the dipole (physical or induced).
The strength of the polarization is described by the dipole moment of the dipole, $\vec{\mathbf{p}}=q\vec{\mathbf{d}}$ .

Answers to Check Your Understanding

1.1 The force would point outward.

1.2 The net force would point $58^{\circ}$ below the $-x$ -axis.

1.3 $\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$

1.4 We will no longer be able to take advantage of symmetry. Instead, we will need to calculate each of the two components of the electric field with their own integral.

1.5 The point charge would be $Q=\sigma ab$ where $a$ and $b$ are the sides of the rectangle but otherwise identical.

1.6 The electric field would be zero in between, and have magnitude $\frac{\sigma}{\epsilon_0}$ everywhere else.

Conceptual Questions

1.1 Electric Charge

1. There are very large numbers of charged particles in most objects. Why, then, don’t most objects exhibit static electricity?

2. Why do most objects tend to contain nearly equal numbers of positive and negative charges?

3. A positively charged rod attracts a small piece of cork. (a) Can we conclude that the cork is negatively charged? (b) The rod repels another small piece of cork. Can we conclude that this piece is positively charged?

4. Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.

5. How would you determine whether the charge on a particular rod is positive or negative?

1.2 Conductors, Insulators, and Charging by Induction

6. An eccentric inventor attempts to levitate a cork ball by wrapping it with foil and placing a large negative charge on the ball and then putting a large positive charge on the ceiling of his workshop. Instead, while attempting to place a large negative charge on the ball, the foil flies off. Explain.

7. When a glass rod is rubbed with silk, it becomes positive and the silk becomes negative—yet both attract dust. Does the dust have a third type of charge that is attracted to both positive and negative? Explain.

8. Why does a car always attract dust right after it is polished? (Note that car wax and car tires are insulators.)

9. Does the uncharged conductor shown below experience a net electric force?

A sphere is shown suspended by a thread from the ceiling. A negatively charged rod is brought near the sphere.

10. While walking on a rug, a person frequently becomes charged because of the rubbing between his shoes and the rug. This charge then causes a spark and a slight shock when the person gets close to a metal object. Why are these shocks so much more common on a dry day?

11. Compare charging by conduction to charging by induction.

12. Small pieces of tissue are attracted to a charged comb. Soon after sticking to the comb, the pieces of tissue are repelled from it. Explain.

13. Trucks that carry gasoline often have chains dangling from their undercarriages and brushing the ground. Why?

14. Why do electrostatic experiments work so poorly in humid weather?

15. Why do some clothes cling together after being removed from the clothes dryer? Does this happen if they’re still damp?

16. Can induction be used to produce charge on an insulator?

17. Suppose someone tells you that rubbing quartz with cotton cloth produces a third kind of charge on the quartz. Describe what you might do to test this claim.

18. A handheld copper rod does not acquire a charge when you rub it with a cloth. Explain why.

19. Suppose you place a charge $q$ near a large metal plate. (a) If $q$ is attracted to the plate, is the plate necessarily charged? (b) If $q$ is repelled by the plate, is the plate necessarily charged?

1.3 Coulomb’s Law

20. Would defining the charge on an electron to be positive have any effect on Coulomb’s law?

21. An atomic nucleus contains positively charged protons and uncharged neutrons. Since nuclei do stay together, what must we conclude about the forces between these nuclear particles?

22. Is the force between two fixed charges influenced by the presence of other charges?

1.4 Electric Field

23. When measuring an electric field, could we use a negative rather than a positive test charge?

24. During fair weather, the electric field due to the net charge on Earth points downward. Is Earth charged positively or negatively?

25. If the electric field at a point on the line between two charges is zero, what do you know about the charges?

26. Two charges lie along the $x$ -axis. Is it true that the net electric field always vanishes at some point (other than infinity) along the $x$ -axis?

1.5 Calculating Electric Fields of Charge Distributions

27. Give a plausible argument as to why the electric field outside an infinite charged sheet is constant.

28. Compare the electric fields of an infinite sheet of charge, an infinite, charged conducting plate, and infinite, oppositely charged parallel plates.

29. Describe the electric fields of an infinite charged plate and of two infinite, charged parallel plates in terms of the electric field of an infinite sheet of charge.

30. A negative charge is placed at the center of a ring of uniform positive charge. What is the motion (if any) of the charge? What if the charge were placed at a point on the axis of the ring other than the center?

1.6 Electric Field Lines

31. If a point charge is released from rest in a uniform electric field, will it follow a field line? Will it do so if the electric field is not uniform?

32. Under what conditions, if any, will the trajectory of a charged particle not follow a field line?

33. How would you experimentally distinguish an electric field from a gravitational field?

34. A representation of an electric field shows 10 field lines perpendicular to a square plate. How many field lines should pass perpendicularly through the plate to depict a field with twice the magnitude?

35. What is the ratio of the number of electric field lines leaving a charge 10 $q$ and a charge $q$ ?

1.7 Electric Dipoles

36. What are the stable orientation(s) for a dipole in an external electric field? What happens if the dipole is slightly perturbed from these orientations?

Problems

1.1 Electric Charge

37. Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of $−2.00~\mathrm{nC}$ ? (b) How many electrons must be removed from a neutral object to leave a net charge of $0.500~\mu\mathrm{C}$ ?

38. If $1.80\times10^{20}$ electrons move through a pocket calculator during a full day’s operation, how many coulombs of charge moved through it?

39. To start a car engine, the car battery moves $3.75\times10^{21}$ electrons through the starter motor. How many coulombs of charge were moved?

40. A certain lightning bolt moves $40.0~\mathrm{C}$ of charge. How many fundamental units of charge is this?

41. A $2.5$ – $\mathrm{g}$ copper penny is given a charge of $-2.0\times10^{-9}~\mathrm{C}$ . (a) How many excess electrons are on the penny? (b) By what percent do the excess electrons change the mass of the penny?

42. A $2.5$ – $\mathrm{g}$ copper penny is given a charge of $4.0\times10^{-9}~\mathrm{C}$ . (a) How many electrons are removed from the penny? (b) If no more than one electron is removed from an atom, what percent of the atoms are ionized by this charging process?

1.2 Conductors, Insulators, and Charging by Induction

43. Suppose a speck of dust in an electrostatic precipitator has $1.0000\times10^{12}$ protons in it and has a net charge of $−5.00~\mathrm{nC}$ (a very large charge for a small speck). How many electrons does it have?

44. An amoeba has $1.00\times{10}^{16}$ protons and a net charge of $0.300~\mathrm{pC}$ . (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons?

45. A $50.0$ – $\mathrm{g}$ ball of copper has a net charge of $2.00~\mu\mathrm{C}$ . What fraction of the copper’s electrons has been removed? (Each copper atom has $29$ protons, and copper has an atomic mass of $63.5~\mathrm{u}$ .)

46. What net charge would you place on a $100$ – $\mathrm{g}$ piece of sulfur if you put an extra electron on $1$ in $10^{12}$ of its atoms? (Sulfur has an atomic mass of $32.1~\mathrm{u}$ .)

47. How many coulombs of positive charge are there in $4.00\mathrm{kg}$ of plutonium, given its atomic mass is $244$ and that each plutonium atom has $94$ protons?

1.3 Coulomb’s Law

48. Two point particles with charges $+3~\mu\mathrm{C}$ and $+5~\mu\mathrm{C}$ are held in place by $3$ – $\mathrm{N}$ forces on each charge in appropriate directions. (a) Draw a free-body diagram for each particle. (b) Find the distance between the charges.

49. Two charges $+3~\mu\mathrm{C}$ and $+12~\mu\mathrm{C}$ are fixed $1~\mathrm{m}$ apart, with the second one to the right. Find the magnitude and direction of the net force on a $-2$ – $\mathrm{nC}$ charge when placed at the following locations: (a) halfway between the two (b) half a meter to the left of the $+3~\mu\mathrm{C}$ charge (c) half a meter above the $+12~\mu\mathrm{C}$ charge in a direction perpendicular to the line joining the two fixed charges.

50. In a salt crystal, the distance between adjacent sodium and chloride ions is $2.82\times10^{-10}~\mathrm{m}$ What is the force of attraction between the two singly charged ions?

51. Protons in an atomic nucleus are typically $10^{-15}~\mathrm{m}$ apart. What is the electric force of repulsion between nuclear protons?

52. Suppose Earth and the Moon each carried a net negative charge $-Q$ . Approximate both bodies as point masses and point charges. (a) What value of $Q$ is required to balance the gravitational attraction between Earth and the Moon? (b) Does the distance between Earth and the Moon affect your answer? Explain. (c) How many electrons would be needed to produce this charge?

53. Point charges $q_1=50~\mu\mathrm{C}$ and $q_2=-25~\mu\mathrm{C}$ are placed $1.0~\mathrm{m}$ apart. What is the force on a third charge $q_3=10~\mu\mathrm{C}$ placed midway between $q_1$ and $q_2$ ?

54. Where must $q_3$ of the preceding problem be placed so that the net force on it is zero?

55. Two small balls, each of mass $5.0~\mathrm{g}$ , are attached to silk threads $50\mathrm{cm}$ long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge $Q$ , the threads hang at $5.0^{\circ}$ to the vertical, as shown below. What is the magnitude of $Q$ ? What are the signs of the two charges?

Two small balls are attached to threads which are in turn tied to the same point on the ceiling. The threads hang at an angle of 5.0 degrees to either side of the vertical. Each ball has a charge Q.

56. Point charges $Q_1=2.0~\mu\mathrm{C}$ and $Q_2=4.0~\mu\mathrm{C}$ are located at $\vec{\mathbf{r}}_1=(4.0\hat{\mathbf{i}}-2.0\hat{\mathbf{j}}+5.0\hat{\mathbf{k}})~\mathrm{m}$ and $\vec{\mathbf{r}}_2=(8.0\hat{\mathbf{i}}+5.0\hat{\mathbf{j}}-9.0\hat{\mathbf{k}})~\mathrm{m}$ . What is the force of $Q_2$ on $Q_1$ ?

57. The net excess charge on two small spheres (small enough to be treated as point charges) is $Q$ . Show that the force of repulsion between the spheres is greatest when each sphere has an excess charge $Q/2$ . Assume that the distance between the spheres is so large compared with their radii that the spheres can be treated as point charges.

58. Two small, identical conducting spheres repel each other with a force of $0.050~\mathrm{N}$ when they are $0.25~\mathrm{m}$ apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of $0.060~\mathrm{N}$ . What is the original charge on each sphere?

59. A charge $q=2.0~\mu\mathrm{C}$ is placed at the point $P$ shown below. What is the force on $q$ ?

Two charges are shown, placed on a horizontal line and separated by 2.0 meters. The charge on the left is a positive 1.0 micro Coulomb charge. The charge on the right is a negative 2.0 micro Coulomb charge. Point P is 1.0 to the right of the negative charge.

60. What is the net electric force on the charge located at the lower right-hand corner of the triangle shown here?

Charges are shown at the vertices of an equilateral triangle with sides length a. The bottom of the triangle is on the x axis of an x y coordinate system, and the bottom left vertex is at the origin. The charge at the origin is positive q. The charge at the bottom right hand corner is also positive q. The charge at the top vertex is negative two q.

61. Two fixed particles, each of charge $5.0\times10^{-6}~\mathrm{C}$ are $24~\mathrm{cm}$ apart. What force do they exert on a third particle of charge $-2.5\times10^{-6}~\mathrm{C}$ that is $13~\mathrm{cm}$ from each of them?

62. The charges $q_1=2.0\times10^{-7}~\mathrm{C}$ , $q_2=-4.0\times10^{-7}~\mathrm{C}$ , and $q_3=-1.0\times10^{-7}~\mathrm{C}$ are placed at the corners of the triangle shown below. What is the force on $q_1$ ?

Charges are shown at the vertices of a right triangle. The bottom of the triangle is length 4 meters, the vertical side on the left is length 3 meters, and the hypotenuse is length 5 meters. The charge at the top is q sub one and positive, the charge at the bottom left is q sub 3 and negative and the charge at the bottom right is q sub 2 and negative.

63. What is the force on the charge $q$ at the lower-right-hand corner of the square shown here?

Charges are shown at the corners of a square with sides length a. All of the charges are positive and all are magnitude q.

64. Point charges $q_1=10~\mu\mathrm{C}$ and $q_2=-30~\mu\mathrm{C}$ are fixed at $\vec{\mathbf{r}}_1=(3.0\hat{\mathbf{i}}-4.0\hat{\mathbf{j}})~\mathrm{m}$ and $\vec{\mathbf{r}}_1=(9.0\hat{\mathbf{i}}+6.0\hat{\mathbf{j}})~\mathrm{m}$ . What is the force of $q_2$ on $q_1$ ?

1.4 Electric Field

65. A particle of charge $2.0\times10^{-8}~\mathrm{C}$ experiences an upward force of magnitude $4.0\times10^{-6}~\mathrm{N}$ when it is placed in a particular point in an electric field. (a) What is the electric field at that point? (b) If a charge $q=-1.0\times10^{-8}~\mathrm{C}$ is placed there, what is the force on it?

66. On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately $100~\mathrm{N/C}$ . Compare the gravitational and electric forces on a small dust particle of mass $2.0\times10^{-15}~\mathrm{g}$ that carries a single electron charge. What is the acceleration (both magnitude and direction) of the dust particle?

67. Consider an electron that is $10^{-10}~\mathrm{m}$ from an alpha particle ( $q=3.2\times10^{-19}~\mathrm{C}$ ). (a) What is the electric field due to the alpha particle at the location of the electron? (b) What is the electric field due to the electron at the location of the alpha particle? (c) What is the electric force on the alpha particle? On the electron?

68. Each the balls shown below carries a charge $q$ and has a mass $m$ . The length of each thread is $l$ , and at equilibrium, the balls are separated by an angle $2\theta$ . How does $\theta$ vary with $q$ and $l$ ? Show that $\theta$ satisfies $\sin^{2}\theta\tan\theta=\frac{q^2}{16\pi\epsilon_0gl^2m}$ .

Two small balls are attached to threads of length l which are in turn tied to the same point on the ceiling. The threads hang at an angle of theta to either side of the vertical. Each ball has a charge q and mass m.

69. What is the electric field at a point where the force on a charge $q=-2.0\times10^{-6}~\mathrm{C}$ is $\left(4.0\hat{\mathbf{i}}-6.0\hat{\mathbf{j}}\right)\times10^{-6}~\mathrm{N}$ ?

70. A proton is suspended in the air by an electric field at the surface of Earth. What is the strength of this electric field?

71. The electric field in a particular thundercloud is $2.0\times10^{5}~\mathrm{N/C}$ What is the acceleration of an electron in this field?

72. A small piece of cork whose mass is $2.0~\mathrm{g}$ is given a charge of $5.0\times10^{-7}~\mathrm{C}$ . What electric field is needed to place the cork in equilibrium under the combined electric and gravitational forces?

73. If the electric field is $100~\mathrm{N/C}$ at a distance of $50~\mathrm{cm}$ from a point charge $q$ , what is the value of $q$ ?

74. What is the electric field of a proton at the first Bohr orbit for hydrogen ( $r=5.29\times10^{-11}~\mathrm{m}$ )? What is the force on the electron in that orbit?

75. (a) What is the electric field of an oxygen nucleus at a point that is $10^{-10}~\mathrm{m}$ from the nucleus? (b) What is the force this electric field exerts on a second oxygen nucleus placed at that point?

76. Two point charges, $q_1=2.0\times10^{-7}~\mathrm{C}$ and $q_2=-6.0\times10^{-8}~\mathrm{C}$ are held $25.0~\mathrm{cm}$ apart. (a) What is the electric field at a point $5.0~\mathrm{cm}$ from the negative charge and along the line between the two charges? (b)What is the force on an electron placed at that point?

77. Point charges $q_1=50~\mu\mathrm{C}$ and $q_1=-25~\mu\mathrm{C}$ are placed $1.0~\mathrm{m}$ apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge $q_3=20~\mu\mathrm{C}$ situated there?

78. Can you arrange the two point charges $q_1=-2.0\times10^{-6}~\mathrm{C}$ and $q_2=4.0\times10^{-6}~\mathrm{C}$ along the $x$ -axis so that $E=0$ at the origin?

79. Point charges $q_1=q_2=4.0\times10^{-6}~\mathrm{C}$ are fixed on the $x$ -axis at $x=-3.0~\mathrm{m}$ and $x=3.0~\mathrm{m}$ What charge $q$ must be placed at the origin so that the electric field vanishes at $x=0$ , $y=3.0~\mathrm{m}$

1.5 Calculating Electric Fields of Charge Distributions

80. A thin conducting plate $1.0~\mathrm{m}$ on the side is given a charge of $-2.0\times10^{-6}~\mathrm{C}$ . An electron is placed $1.0~\mathrm{cm}$ above the centre of the plate. What is the acceleration of the electron?

81. Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at $\lambda=4.0\times10^{-6}~\mathrm{C/m}$

82. Two thin conducting plates, each $25.0~\mathrm{cm}$ on a side, are situated parallel to one another and $5.0~\mathrm{mm}$ apart. If $10^{11}$ electrons are moved from one plate to the other, what is the electric field between the plates?

83. The charge per unit length on the thin rod shown below is $\lambda$ . What is the electric field at the point $P$ ? (Hint: Solve this problem by first considering the electric field $d\vec{\mathbf{E}}$ at $P$ due to a small segment $dx$ of the rod, which contains charge $dq=\lambda dx$ . Then find the net field by integrating $d\vec{\mathbf{E}}$ over the length of the rod.)

A horizontal rod of length L is shown. The rod has total charge q. Point P is a distance a to the right of the right end of the rod.

84. The charge per unit length on the thin semicircular wire shown below is $\lambda$ . What is the electric field at the point $P$ ?

A semicircular arc of radius r is shown. The arc has total charge q. Point P is at the center of the circle of which the arc is a part.

85. Two thin parallel conducting plates are placed $2.0~\mathrm{cm}$ apart. Each plate is $2.0~\mathrm{cm}$ on a side; one plate carries a net charge of $8.0~\mu\mathrm{C},$ and the other plate carries a net charge of $-8.0~\mu\mathrm{C}.$ What is the charge density on the inside surface of each plate? What is the electric field between the plates?

86. A thin conducing plate $2.0~\mathrm{m}$ on a side is given a total charge of $-10.0~\mu\mathrm{C}.$ (a) What is the electric field $1.0~\mathrm{cm}$ above the plate? (b) What is the force on an electron at this point? (c) Repeat these calculations for a point $2.0~\mathrm{cm}$ above the plate. (d) When the electron moves from $1.0$ to $2.0~\mathrm{cm}$ above the plate, how much work is done on it by the electric field?

87. A total charge $q$ is distributed uniformly along a thin, straight rod of length $L$ (see below). What is the electric field at $P_1$ ? At $P_2$ ?
A horizontal rod of length L is shown. The rod has total charge q. Point P 1 is a distance a over 2 above the midpoint of the rod, so that the horizontal distance from P 1 to each end of the rod is L over 2. Point P 2 is a distance a to the right of the right end of the rod.

88. Charge is distributed along the entire $x$ -axis with uniform density $\lambda$ How much work does the electric field of this charge distribution do on an electron that moves along the $y$ -axis from $y=a$ to $y=b$ ?

89. Charge is distributed along the entire $x$ -axis with uniform density $\lambda_x$ and along the entire $y$ -axis with uniform density $\lambda_y$ Calculate the resulting electric field at (a) $\vec{\mathbf{r}}=a\hat{\mathbf{i}}+b\hat{\mathbf{j}}$ and (b) $\vec{\mathbf{r}}=c\hat{\mathbf{k}}$

90. A rod bent into the arc of a circle subtends an angle $2\theta$ at the centre $P$ of the circle (see below). If the rod is charged uniformly with a total charge $Q$ , what is the electric field at $P$ ?

An arc that is part of a circle of radius R and with center P is shown. The arc extends from an angle theta to the left of vertical to an angle theta to the right of vertical.

91. A proton moves in the electric field $\vec{\mathbf{E}}=100\hat{\mathbf{i}}~\mathrm{N/C}$ . (a) What are the force on and the acceleration of the proton? (b) Do the same calculation for an electron moving in this field.

92. An electron and a proton, each starting from rest, are accelerated by the same uniform electric field of $200~\mathrm{N/C}$ . Determine the distance and time for each particle to acquire a kinetic energy of $3.2\times10^{-16}~\mathrm{J}$

93. A spherical water droplet of radius $25~\mu\mathrm{m}$ carries an excess $250$ electrons. What vertical electric field is needed to balance the gravitational force on the droplet at the surface of the earth?

94. A proton enters the uniform electric field produced by the two charged plates shown below. The magnitude of the electric field is $4.0\times10^{5}~\mathrm{N/C},$ and the speed of the proton when it enters is $1.5\times10^7~\mathrm{m/s}.$ What distance $d$ has the proton been deflected downward when it leaves the plates?

Two oppositely charged horizontal plates are parallel to each other. The upper plate is positive and the lower is negative. The plates are 12.0 centimeters long. The path of a positive proton is shown passing from left to right between the plates. It enters moving horizontally and deflects down toward the negative plate, emerging a distance d below the straight line trajectory.

95. Shown below is a small sphere of mass $0.25~\mathrm{g}$ that carries a charge of $9.0\times10^{-10}~\mathrm{C}.$ The sphere is attached to one end of a very thin silk string $5.0~\mathrm{cm}$ long. The other end of the string is attached to a large vertical conducting plate that has a charge density of $30\times10^{-6}~\mathrm{C/m}^2.$ What is the angle that the string makes with the vertical?

A small sphere is attached to the lower end of a string. The other end of the string is attached to a large vertical conducting plate that has a uniform positive charge density. The string makes an angle of theta with the vertical.

96. Two infinite rods, each carrying a uniform charge density $\lambda,$ are parallel to one another and perpendicular to the plane of the page. (See below.) What is the electrical field at $P_1$ ? At $P_2$ ?

An end view of the arrangement in the problem is shown. Two rods are parallel to one another and perpendicular to the plane of the page. They are separated by a horizontal distance of a. Pint P 1 is a distance of a over 2 above the midpoint between the rods, and so also a distance of a over 2 horizontally from each rod. Point P 2 is a distance of a to the right of the rightmost rod.

97. Positive charge is distributed with a uniform density $\lambda$ along the positive $x$ -axis from $r$ to $\infty$ , along the positive $y$ -axis from $r$ to $\infty$ , and along a $90^{\circ}$ arc of a circle of radius $r,$ as shown below. What is the electric field at $O$ ?

A uniform distribution of positive charges is shown on an x y coordinate system. The charges are distributed along a 90 degree arc of a circle of radius r in the first quadrant, centered on the origin. The distribution continues along the positive x and y axes from r to infinity.

98. From a distance of $10\mathrm{cm}$ , a proton is projected with a speed of $v=4.0\times10^{6}~\mathrm{m/s}$ directly at a large, positively charged plate whose charge density is $\sigma=2.0\times10^{-5}~\mathrm{C/m}^2$ . (See below.) (a) Does the proton reach the plate? (b) If not, how far from the plate does it turn around?

A positive charge is shown at a distance of 10 centimeters and moving to the right with a speed of 4.0 times 10 to the 6 meters per second, directly toward a large, positively and uniformly charged vertical plate.

99. A particle of mass $m$ and charge $-q$ moves along a straight line away from a fixed particle of charge $Q$ . When the distance between the two particles is $r_0,$ $-q$ is moving with a speed $v_0$ . (a) Use the work-energy theorem to calculate the maximum separation of the charges. (b) What do you have to assume about v0 to make this calculation? (c) What is the minimum value of $v_0$ such that $-q$ escapes from $Q$ ?

1.6 Electric Field Lines

100. Which of the following electric field lines are incorrect for point charges? Explain why.

101. In this exercise, you will practice drawing electric field lines. Make sure you represent both the magnitude and direction of the electric field adequately. Note that the number of lines into or out of charges is proportional to the charges. (a) Draw the electric field lines map for two charges $+20~\mu\mathrm{C}$ and $-20~\mu\mathrm{C}$ situated $5~\mathrm{cm}$ from each other. (b) Draw the electric field lines map for two charges $+20~\mu\mathrm{C}$ and $+20~\mu\mathrm{C}$ situated $5~\mathrm{cm}$ from each other. (c) Draw the electric field lines map for two charges $+20~\mu\mathrm{C}$ and $-30~\mu\mathrm{C}$ situated $5~\mathrm{cm}$ from each other.

102. Draw the electric field for a system of three particles of charges $+1~\mu\mathrm{C},$ $+2~\mu\mathrm{C},$ and $-3~\mu\mathrm{C}$ fixed at the corners of an equilateral triangle of side $2~\mathrm{cm}$ .

103. Two charges of equal magnitude but opposite sign make up an electric dipole. A quadrupole consists of two electric dipoles are placed anti-parallel at two edges of a square as shown.

Four charges are shown at the corners of a square. At the top left is positive 10 nano Coulombs. At the top right is negative 10 nano Coulombs. At the bottom left is negative 10 nano Coulombs. At the bottom right is positive 10 nano Coulombs.

Draw the electric field of the charge distribution.

104. Suppose the electric field of an isolated point charge decreased with distance as $1/r^{2+\delta}$ rather than as $1/r^2$ . Show that it is then impossible to draw continuous field lines so that their number per unit area is proportional to $E$ .

1.7 Electric Dipoles

105. Consider the equal and opposite charges shown below. (a) Show that at all points on the $x$ -axis for which $|x|\gg a$ , $E\approx Qa/2\pi\epsilon_0x^3$ . (b) Show that at all points on the $y$ -axis for which $|y|\gg a$ , $E\approx Qa/\pi\epsilon_0y^3$ .
Two charges are shown on the y axis of an x y coordinate system. Charge +Q is a distance a above the origin, and charge −Q is a distance a below the origin.

Two charges are shown on the y axis of an x y coordinate system. Charge +Q is a distance a above the origin, and charge −Q is a distance a below the origin.

106. (a) What is the dipole moment of the configuration shown above? If $Q=4.0~\mu\mathrm{C}$ , (b) what is the torque on this dipole with an electric field of $4.0\times10^5~\mathrm{N/C}\hat{\mathbf{i}}$ ? (c) What is the torque on this dipole with an electric field of $-4.0\times10^5~\mathrm{N/C}\hat{\mathbf{i}}$ ? (d) What is the torque on this dipole with an electric field of $\pm4.0\times10^5~\mathrm{N/C}\hat{\mathbf{j}}$ ?

107. A water molecule consists of two hydrogen atoms bonded with one oxygen atom. The bond angle between the two hydrogen atoms is $104^{\circ}$ (see below). Calculate the net dipole moment of a water molecule that is placed in a uniform, horizontal electric field of magnitude $2.3\times10^{-8}~\mathrm{N/C}$ . (You are missing some information for solving this problem; you will need to determine what information you need, and look it up.)

A schematic representation of the outer electron cloud of a neutral water molecule is shown. Three atoms are at the vertices of a triangle. The hydrogen atom has positive q charge and the oxygen atom has minus two q charge, and the angle between the line joining each hydrogen atom with the oxygen atom is one hundred and four degrees. The cloud density is shown as being greater at the oxygen atom.

Additional Problems

108. Point charges $q_1=2.0~\mu\mathrm{C}$ and $q_2=4.0~\mu\mathrm{C}$ are located at $\vec{\mathbf{r}}_1=\left(4.0\hat{\mathbf{i}}-2.0\hat{\mathbf{j}}+2.0\hat{\mathbf{j}}\right)~\mathrm{m}$ and $\vec{\mathbf{r}}_2=\left(4.0\hat{\mathbf{i}}-2.0\hat{\mathbf{j}}+2.0\hat{\mathbf{j}}\right)~\mathrm{m}$ . What is the force of $q_2$ on $q_1$ ?

109. What is the force on the $5.0$ – $\mu\mathrm{C}$ charge shown below?

The following charges are shown on an x y coordinate system: Minus 3.0 micro Coulomb on the x axis, 3.0 meters to the left of the origin. Positive 5.0 micro Coulomb at the origin. Positive 9.0 micro Coulomb on the x axis, 3.0 meters to the right of the origin. Positive 6.0 micro Coulomb on the y axis, 3.0 meters above the origin.

110. What is the force on the $2.0$ – $\mu\mathrm{C}$ charge placed at the centre of the square shown below?

Charges are shown at the corners of a square with sides length 1 meter. The top left charge is positive 5.0 micro Coulombs. The top right charge is positive 4.0 micro Coulombs. The bottom left charge is negative 4.0 micro Coulombs. The bottom right charge is positive 2.0 micro Coulombs. A fifth charge of positive 2.0 micro Coulombs is at the center of the square.

111. Four charged particles are positioned at the corners of a parallelogram as shown below. If $q=5.0~\mu\mathrm{C}$ and $Q=8.0~\mu\mathrm{C},$ what is the net force on $q$ ?

112. A charge $Q$ is fixed at the origin and a second charge $q$ moves along the $x$ -axis, as shown below. How much work is done on $q$ by the electric force when $q$ moves from $x_1$ to $x_2?$

A charge Q is shown at the origin and a second charge q is shown to its right, on the x axis, moving to the right. Both are positive charges. Point x 1 is between the charges. Point x 2 is to the right of both.

113. A charge $q=-2.0~\mu\mathrm{C}$ is released from rest when it is $2.0~\mathrm{m}$ from a fixed charge $Q=6.0~\mu\mathrm{C}.$ What is the kinetic energy of $q$ when it is $1.0~\mathrm{m}$ from $Q$ ?

114. What is the electric field at the midpoint $M$ of the hypotenuse of the triangle shown below?

Charges are shown at the vertices of an isosceles right triangle whose sides are length a and those hypotenuse is length M. The right angle is the bottom right corner. The charge at the right angle is positive 2 q. Both of the other two charges are positive q.

115. Find the electric field at $P$ for the charge configurations shown below.

116. (a) What is the electric field at the lower-right-hand corner of the square shown below? (b) What is the force on a charge $q$ placed at that point?

A square with sides of length a is shown. Three charges are shown as follows: At the top left, a charge of negative 2 q. At the top right, a charge of positive q. At the lower left, a charge of positive q.

117. Point charges are placed at the four corners of a rectangle as shown below: $q_1=2.0\times10^{-6}~\mathrm{C},$ $q_2=-2.0\times10^{-6}~\mathrm{C},$ $q_3=4.0\times10^{-6}~\mathrm{C},$ and $q_4=1.0\times10^{-6}~\mathrm{C}.$ What is the electric field at $P$ ?

118. Three charges are positioned at the corners of a parallelogram as shown below. (a) If $Q=8.0~\mu\mathrm{C},$ what is the electric field at the unoccupied corner? (b) What is the force on a $5.0$ – $\mu\mathrm{C}$ charge placed at this corner?

Three charges are positioned at the corners of a parallelogram. The top and bottom of the parallelogram are horizontal and are 3.0 meters long. The sides are at a thirty degree angle to the x axis. The vertical height of the parallelogram is 1.0 meter. The charges are a positive Q in the lower left corner, positive 2 Q in the lower right corner, and negative 3 Q in the upper left corner.

119. A positive charge $q$ is released from rest at the origin of a rectangular coordinate system and moves under the influence of the electric field $\vec{\mathrm{E}}=E_0(1+x/a)\hat{\mathbf{i}}.$ What is the kinetic energy of $q$ when it passes through $x=3a$ ?

120. A particle of charge $-q$ and mass $m$ is placed at the centre of a uniformly charged ring of total charge $Q$ and radius $R$ . The particle is displaced a small distance along the axis perpendicular to the plane of the ring and released. Assuming that the particle is constrained to move along the axis, show that the particle oscillates in simple harmonic motion with a frequency $f=\frac{1}{2\pi}\sqrt{\frac{qQ}{4\pi\epsilon_0mR^3}}.$

121. Charge is distributed uniformly along the entire $y$ -axis with a density $\lambda_y$ and along the positive $x$ -axis from $x=a$ to $x=b$ with a density $\lambda_x$ What is the force between the two distributions?

122. The circular arc shown below carries a charge per unit length $\lambda=\lambda_0\cos\theta,$ where $\theta$ is measured from the $x$ -axis. What is the electric field at the origin?

An arc that is part of a circle of radius r and with center at the origin of an x y coordinate system is shown. The arc extends from an angle theta sub zero above the x axis to an angle theta sub zero below the x axis.

123. Calculate the electric field due to a uniformly charged rod of length $L$ , aligned with the $x$ -axis with one end at the origin; at a point $P$ on the $z$ -axis.

124. The charge per unit length on the thin rod shown below is $\lambda.$ What is the electric force on the point charge $q$ ? Solve this problem by first considering the electric force $d\vec{\mathbf{F}}$ on $q$ due to a small segment $dx$ of the rod, which contains charge $\lambda dx.$ Then, find the net force by integrating $d\vec{\mathbf{F}}$ over the length of the rod.

A rod of length l is shown. The rod lies on the horizontal axis, with its left end at the origin. A positive charge q is on the x axis, a distance a to the right of the right end of the rod.

125. The charge per unit length on the thin rod shown here is $\lambda.$ What is the electric force on the point charge $q$ ? (See the preceding problem.)

A rod of length l is shown. The rod lies on the horizontal axis, with its center at the origin, so the ends are a distance of l over 2 to the left and right of the origin. A positive charge q is on the y axis, a distance a to above the origin.

126. The charge per unit length on the thin semicircular wire shown below is $\lambda.$ What is the electric force on the point charge $q$ ? (See the preceding problems.)

A semicircular arc that the upper half of a circle of radius R is shown. A positive charge q is at the center of the circle.

10

2 Gauss's Law

Chapter Outline

2.1 Electric Flux

2.2 Explaining Gauss’s Law

2.3 Applying Gauss’s Law

2.4 Conductors in Electrostatic Equilibrium

Chapter 2 Review

Figure 2.0.1 This chapter introduces the concept of flux, which relates a physical quantity and the area through which it is flowing. Although we introduce this concept with the electric field, the concept may be used for many other quantities, such as fluid flow. (credit: modification of work by “Alessandro”/Flickr)

Flux is a general and broadly applicable concept in physics. However, in this chapter, we concentrate on the flux of the electric field. This allows us to introduce Gauss’s law, which is particularly useful for finding the electric fields of charge distributions exhibiting spatial symmetry. The main topics discussed here are

Electric flux. We define electric flux for both open and closed surfaces.
Gauss’s law. We derive Gauss’s law for an arbitrary charge distribution and examine the role of electric flux in Gauss’s law.
Calculating electric fields with Gauss’s law. The main focus of this chapter is to explain how to use Gauss’s law to find the electric fields of spatially symmetrical charge distributions. We discuss the importance of choosing a Gaussian surface and provide examples involving the applications of Gauss’s law.
Electric fields in conductors. Gauss’s law provides useful insight into the absence of electric fields in conducting materials.

So far, we have found that the electrostatic field begins and ends at point charges and that the field of a point charge varies inversely with the square of the distance from that charge. These characteristics of the electrostatic field lead to an important mathematical relationship known as Gauss’s law. This law is named in honor of the extraordinary German mathematician and scientist Karl Friedrich Gauss (Figure 2.0.2 ). Gauss’s law gives us an elegantly simple way of finding the electric field, and, as you will see, it can be much easier to use than the integration method described in the previous chapter. However, there is a catch—Gauss’s law has a limitation in that, while always true, it can be readily applied only for charge distributions with certain symmetries.

(Figure 2.0.2) $\begin{gather*}.\end{gather*}$

Figure 2.0.2 Karl Friedrich Gauss (1777–1855) was a legendary mathematician of the nineteenth century. Although his major contributions were to the field of mathematics, he also did important work in physics and astronomy.

11

2.1 Electric Flux

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Define the concept of flux
Describe electric flux
Calculate electric flux for a given situation

The concept of flux describes how much of something goes through a given area. More formally, it is the dot product of a vector field (in this chapter, the electric field) with an area. You may conceptualize the flux of an electric field as a measure of the number of electric field lines passing through an area (Figure 2.1.1). The larger the area, the more field lines go through it and, hence, the greater the flux; similarly, the stronger the electric field is (represented by a greater density of lines), the greater the flux. On the other hand, if the area rotated so that the plane is aligned with the field lines, none will pass through and there will be no flux.

(Figure 2.1.1) $\begin{gather*}.\end{gather*}$

Figure shows a shaded area in the center. Several arrows pointing right are shown behind, in front of and passing through the shaded area. These are labeled electric field.

Figure 2.1.1 The flux of an electric field through the shaded area captures information about the “number” of electric field lines passing through the area. The numerical value of the electric flux depends on the magnitudes of the electric field and the area, as well as the relative orientation of the area with respect to the direction of the electric field.

A macroscopic analogy that might help you imagine this is to put a hula hoop in a flowing river. As you change the angle of the hoop relative to the direction of the current, more or less of the flow will go through the hoop. Similarly, the amount of flow through the hoop depends on the strength of the current and the size of the hoop. Again, flux is a general concept; we can also use it to describe the amount of sunlight hitting a solar panel or the amount of energy a telescope receives from a distant star, for example.

To quantify this idea, Figure 2.1.2(a) shows a planar surface $S_1$ of area $A_1$ that is perpendicular to the uniform electric field $\vec{\mathbf{E}}=E\hat{\mathbf{y}}$ . If $N$ field lines pass through $S_1$ , then we know from the definition of electric field lines (Electric Charges and Fields) that $N/A_1\propto$ , or $N\propto EA_1$ .

The quantity $EA_1$ is the electric flux through $S_1$ . We represent the electric flux through an open surface like $S_1$ by the symbol $\Phi$ . Electric flux is a scalar quantity and has an SI unit of newton-meters squared per coulomb ( $\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ ). Notice that $N\propto EA_1$ may also be written as $N\propto\Phi$ , demonstrating that electric flux is a measure of the number of field lines crossing a surface.

(Figure 2.1.2) $\begin{gather*}.\end{gather*}$

Figure a shows a rectangular shaded area in the xz plane. This is labeled S1. There are three arrows labeled E passing through S1. They are parallel to the y axis and point along the positive y axis. Figure b, too has plane S1 and arrows E. Another plane, labeled S2 forms an angle theta with plane S1. Their line of intersection is parallel to the x axis. An arrow labeled n hat 2 forms an angle theta with E.

Figure 2.1.2 (a) A planar surface $S_1$ of area $A_1$ is perpendicular to the electric field $E\hat{\mathbf{j}}$ . $N$ field lines cross surface $S_1$ . (b) A surface $S_2$ of area $A_2$ whose projection onto the $xz$ -plane is $S_1$ . The same number of field lines cross each surface.

Now consider a planar surface that is not perpendicular to the field. How would we represent the electric flux? Figure 2.1.2(b) shows a surface $S_2$ of area $A_2$ that is inclined at an angle $\theta$ to the $xz$ -plane and whose projection in that plane is $S_1$ (area $A_1$ ). The areas are related by $A_2\cos\theta=A_1$ . Because the same number of field lines crosses both $S_1$ and $S_2$ , the fluxes through both surfaces must be the same. The flux through $S_2$ is therefore $\Phi=EA_1=EA_2\cos\theta$ . Designating $\hat{\matbf{n}}_2$ as a unit vector normal to $S_2$ (see Figure 2.1.2(b)), we obtain

$\Phi=\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}_2A_2.$

INTERACTIVE

Check out this video to observe what happens to the flux as the area changes in size and angle, or the electric field changes in strength.

Area Vector

For discussing the flux of a vector field, it is helpful to introduce an area vector $\vec{\mathbf{A}}$ . This allows us to write the last equation in a more compact form. What should the magnitude of the area vector be? What should the direction of the area vector be? What are the implications of how you answer the previous question?

The area vector of a flat surface of area $A$ has the following magnitude and direction:

Magnitude is equal to area ( $A$ )
Direction is along the normal to the surface ( $\hat{\mathbf{n}}$ ); that is, perpendicular to the surface.

Since the normal to a flat surface can point in either direction from the surface, the direction of the area vector of an open surface needs to be chosen, as shown in Figure 2.1.3.

(Figure 2.1.3) $\begin{gather*}.\end{gather*}$

Figure shows two horizontal planes labeled A. The first has two arrows pointing up from the plane. The longer is labeled vector A and the shorter is labeled n hat. The second plane has the same two arrows pointing down from the plane.

Figure 2.1.3 The direction of the area vector of an open surface needs to be chosen; it could be either of the two cases displayed here. The area vector of a part of a closed surface is defined to point from the inside of the closed space to the outside. This rule gives a unique direction.

Since $\hat{\mathbf{n}}$ is a unit normal to a surface, it has two possible directions at every point on that surface (Figure 2.1.4(a)). For an open surface, we can use either direction, as long as we are consistent over the entire surface. Part (c) of the figure shows several cases.

(Figure 2.1.4) $\begin{gather*}.\end{gather*}$

Figure a shows a curved rectangular surface. Two arrows originate from a point at its center and point in opposite directions. They are both perpendicular to the surface. They are labeled n hat 1 and n hat 2. Figure b shows a 3 dimensional surface shaped somewhat like a light bulb. There are five arrows labeled n hat, which originate from various points on the surface and point outward, perpendicular to the surface. Figure c shows three rectangular surfaces labeled S1, S2 and S3. Two arrows labeled n hat are perpendicular to S1 and point in opposite directions. Three arrows labeled n hat are perpendicular to S2, one pointing in a direction opposite to the other two. There are three arrows perpendicular to S3. All point outward from the same side of the surface.

Figure 2.1.4 (a) Two potential normal vectors arise at every point on a surface. (b) The outward normal is used to calculate the flux through a closed surface. (c) Only $S_3$ has been given a consistent set of normal vectors that allows us to define the flux through the surface.

However, if a surface is closed, then the surface encloses a volume. In that case, the direction of the normal vector at any point on the surface points from the inside to the outside. On a closed surface such as that of Figure 2.1.4(b), $\hat{\mathbf{n}}$ is chosen to be the outward normal at every point, to be consistent with the sign convention for electric charge.

Electric Flux

Now that we have defined the area vector of a surface, we can define the electric flux of a uniform electric field through a flat area as the scalar product of the electric field and the area vector:

(2.1.1) $\begin{equation*}\Phi=\vec{\mathbf{E}}\cdot\hat{\mathbf{A}}~(\mathrm{uniform}~\vec{\mathbf{E}},~\mathrm{flat\ surface}).\end{equation*}$

Figure 2.1.5 shows the electric field of an oppositely charged, parallel-plate system and an imaginary box between the plates. The electric field between the plates is uniform and points from the positive plate toward the negative plate. A calculation of the flux of this field through various faces of the box shows that the net flux through the box is zero. Why does the flux cancel out here?

(Figure 2.1.5) $\begin{gather*}.\end{gather*}$

A cube ABCDKFGH is shown in the center. A diagonal plane is shown within it from KF to BC. The top surface of the cube, FGHK has a plane labeled minus q slightly above it and parallel to it. Similarly, another plane is labeled plus q is shown slightly below the bottom surface of the cube, parallel to it. Small red arrows are shown pointing upwards from the bottom plane, pointing up to the bottom surface of the cube, pointing up from the top surface of the cube and pointing up to the top plane. These are labeled vector E.

Figure 2.1.5 Electric flux through a cube, placed between two charged plates. Electric flux through the bottom face ( $ABCD$ ) is negative, because $\vec{\mathbf{E}}$ is in the opposite direction to the normal to the surface. The electric flux through the top face ( $FGHK$ ) is positive, because the electric field and the normal are in the same direction. The electric flux through the other faces is zero, since the electric field is perpendicular to the normal vectors of those faces. The net electric flux through the cube is the sum of fluxes through the six faces. Here, the net flux through the cube is equal to zero. The magnitude of the flux through rectangle $BCKF$ is equal to the magnitudes of the flux through both the top and bottom faces.

The reason is that the sources of the electric field are outside the box. Therefore, if any electric field line enters the volume of the box, it must also exit somewhere on the surface because there is no charge inside for the lines to land on. Therefore, quite generally, electric flux through a closed surface is zero if there are no sources of electric field, whether positive or negative charges, inside the enclosed volume. In general, when field lines leave (or “flow out of”) a closed surface, $\Phi$ is positive; when they enter (or “flow into”) the surface, $\Phi$ is negative.

Any smooth, non-flat surface can be replaced by a collection of tiny, approximately flat surfaces, as shown in Figure 2.1.6. If we divide a surface S into small patches, then we notice that, as the patches become smaller, they can be approximated by flat surfaces. This is similar to the way we treat the surface of Earth as locally flat, even though we know that globally, it is approximately spherical.

(Figure 2.1.6) $\begin{gather*}.\end{gather*}$

Figure shows a wavy surface labeled S. Three arrows labeled n hat originate from three different patches on the surface. Longer arrows labeled vector E also originate from each patch. An enlarged view of one patch is shown on the side. It shows n hat to be perpendicular to the patch.

Figure 2.1.6 A surface is divided into patches to find the flux.

To keep track of the patches, we can number them from 1 through $N$ . Now, we define the area vector for each patch as the area of the patch pointed in the direction of the normal. Let us denote the area vector for the $i$ th patch by $\delta\vec{\mathbf{A}}_i$ . (We have used the symbol $\delta$ to remind us that the area is of an arbitrarily small patch.) With sufficiently small patches, we may approximate the electric field over any given patch as uniform. Let us denote the average electric field at the location of the $i$ th patch by $\vec{\mathbf{E}}_i$

$\vec{\mathbf{E}}_i=\mathrm{average~electric~field~over~the~}i\mathrm{th~patch}.$

Therefore, we can write the electric flux $\Phi_i$ through the area of the $i$ th patch as

$\Phi_i=\vec{\mathbf{E}}_i\cdot\delta\vec{\mathbf{A}}_i~(i\mathrm{th~patch}).$

The flux through each of the individual patches can be constructed in this manner and then added to give us an estimate of the net flux through the entire surface $S$ , which we denote simply as $\Phi$ .

$\Phi=\sum_{i=1}^{N}\Phi_i=\sum_{i=1}^{N}\vec{\mathbf{E}}_i\cdot\delta\vec{\mathbf{A}}_i~(N\mathrm{patch~estimate}).$

This estimate of the flux gets better as we decrease the size of the patches. However, when you use smaller patches, you need more of them to cover the same surface. In the limit of infinitesimally small patches, they may be considered to have area $dA$ and unit normal $\hat{\mathbf{n}}$ . Since the elements are infinitesimal, they may be assumed to be planar, and $\vec{\mathbf{E}}_i$ may be taken as constant over any element. Then the flux $d\Phi$ through an area $dA$ is given by $d\Phi=\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA$ . It is positive when the angle between $\vec{\mathbf{E}}_i$ and $\hat{\mathbf{n}}$ is less than $90^{\circ}$ and negative when the angle is greater than $90^{\circ}$ . The net flux is the sum of the infinitesimal flux elements over the entire surface. With infinitesimally small patches, you need infinitely many patches, and the limit of the sum becomes a surface integral. With $\int_{S}$ representing the integral over $S$ ,

(2.1.2) $\begin{equation*}\Phi=\int_{S}\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\int_S\vec{\mathbf{E}}\cdot d\vec{\mathbf{A}}~(\mathrm{open~surface}).\end{equation*}$

In practical terms, surface integrals are computed by taking the antiderivatives of both dimensions defining the area, with the edges of the surface in question being the bounds of the integral.

To distinguish between the flux through an open surface like that of Figure 2.1.2 and the flux through a closed surface (one that completely bounds some volume), we represent flux through a closed surface by

(2.1.3) $\begin{equation*}\Phi=\oint_{S}\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\oint_S\vec{\mathbf{E}}\cdot d\vec{\mathbf{A}}~(\mathrm{closed~surface}).\end{equation*}$

where the circle through the integral symbol simply means that the surface is closed, and we are integrating over the entire thing. If you only integrate over a portion of a closed surface, that means you are treating a subset of it as an open surface.

EXAMPLE 2.1.1

Flux of a Uniform Electric Field

A constant electric field of magnitude $E_0$ points in the direction of the positive $z$ -axis (Figure 2.1.7). What is the electric flux through a rectangle with sides $a$ and $b$ in the (a) $xy$ -plane and in the (b) $xz$ -plane?

(Figure 2.1.7) $\begin{gather*}.\end{gather*}$

A rectangular patch is shown in the xy plane. Its side along the x axis is of length a and its side along the y axis is of length b. Arrows labeled E subscript 0 originate from the plane and point in the positive z direction.

Figure 2.1.7 Calculating the flux of $E_0$ through a rectangular surface.

Strategy

Apply the definition of flux: $\Phi=\vec{\mathbf{E}}\cdot\hat{\mathbf{A}}~(\mathrm{uniform}~\vec{\mathbf{E}})$ , where the definition of dot product is crucial.

Solution

In this case, $\Phi=\vec{\mathbf{E}}_0\cdot\hat{\mathbf{A}}=E_0A=E_0ab$ .
Here, the direction of the area vector is either along the positive $y$ -axis or toward the negative $y$ -axis. Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.

Significance

The relative directions of the electric field and area can cause the flux through the area to be zero.

EXAMPLE 2.1.2

Flux of a Uniform Electric Field through a Closed Surface

A constant electric field of magnitude $E_0$ points in the direction of the positive $z$ -axis (Figure 2.1.8). What is the net electric flux through a cube?

(Figure 2.1.8) $\begin{gather*}.\end{gather*}$

Figure 2.1.8 Calculating the flux of $E_0$ through a closed cubic surface.

Strategy

Apply the definition of flux: $\Phi=\vec{\mathbf{E}}\cdot\hat{\mathbf{A}}~(\mathrm{uniform}~\vec{\mathbf{E}})$ , noting that a closed surface eliminates the ambiguity in the direction of the area vector.

Solution

Through the top face of the cube, $\Phi=\vec{\mathbf{E}}_0\cdot\hat{\mathbf{A}}=E_0A.$

Through the bottom face of the cube, $\Phi=\vec{\mathbf{E}}_0\cdot\hat{\mathbf{A}}=-E_0A.$ because the area vector here points downward.

Along the other four sides, the direction of the area vector is perpendicular to the direction of the electric field. Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.

The net flux is $\Phi_{\mathrm{net}}=E_0A-E_0A+0+0+0+0=0$ .

Significance

The net flux of a uniform electric field through a closed surface is zero.

EXAMPLE 2.1.3

Electric Flux through a Plane, Integral Method

A uniform electric field $\vec{\mathbf{E}}$ of magnitude $10~\mathrm{N/C}$ is directed parallel to the $yz$ -plane at $30^{\circ}$ above the $xy$ -plane, as shown in Figure 2.1.9. What is the electric flux through the plane surface of area $6.0~\mathrm{m}^2$ located in the $xz$ -plane? Assume that $\hat{\mathbf{n}}$ points in the positive $y$ -direction.

(Figure 2.1.9) $\begin{gather*}.\end{gather*}$

A rectangular surface S is shown in the xz plane. Three arrows labeled n hat originate from three points on the surface and point in the positive y direction. Three longer arrows labeled vector E also originate from the same points. They make an angle of 30 degrees with n hat.

Figure 2.1.9 The electric field produces a net electric flux through the surface $S$ .

Strategy

Apply $\Phi=\int_{S}\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA$ , where the direction and magnitude of the electric field are constant.

Solution

The angle between the uniform electric field $\vec{\mathbf{E}}$ and the unit normal $\hat{\mathbf{n}}$ to the planar surface is $30^{\circ}$ . Since both the direction and magnitude are constant, $E$ comes outside the integral. All that is left is a surface integral over $dA$ , which is $A$ . Therefore, using the open-surface equation, we find that the electric flux through the surface is

$\begin{eqnarray*}\Phi&=&\int_{S}\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=EA\cos\theta\\&=&(10~\mathrm{N/C})(6.0~\mathrm{m}^2)(\cos30^{\circ})=52~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}.\end{eqnarray*}$

Significance

Again, the relative directions of the field and the area matter, and the general equation with the integral will simplify to the simple dot product of area and electric field.

CHECK YOUR UNDERSTANDING 2.1

What angle should there be between the electric field and the surface shown in Figure 2.1.9 in the previous example so that no electric flux passes through the surface?

EXAMPLE 2.1.4

Inhomogeneous Electric Field

What is the total flux of the electric field $\vec{\mathbf{E}}=cy^2\hat{\mathbf{k}}$ through the rectangular surface shown in Figure 2.1.10 ?

(Figure 2.1.10) $\begin{gather*}.\end{gather*}$

A rectangle labeled S is shown in the xy plane. Its side along y axis is of length a and that along the x axis measures b. A strip is marked on the rectangle, with its length parallel to x axis. Its length is b and its breadth is dy. Its area is labeled dA equal to b dy. Two arrows are shown perpendicular to S, n hat equal to k hat and vector E equal to cy squared k hat. These point in the positive z direction.

Figure 2.1.10 Since the electric field is not constant over the surface, an integration is necessary to determine the flux.

Strategy

Apply $\Phi=\int_{S}\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA$ . We assume that the unit normal $\hat{\mathbf{n}}$ to the given surface points in the positive $z$ -direction, so $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ . Since the electric field is not uniform over the surface, it is necessary to divide the surface into infinitesimal strips along which $\vec{\mathbf{E}}$ is essentially constant. As shown in Figure 2.1.10, these strips are parallel to the x-axis, and each strip has an area $dA=b\,dy$

Solution

From the open surface integral, we find that the net flux through the rectangular surface is

$\begin{eqnarray*}\Phi&=&\int_{S}\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\int_0^a(cy^2\hat{\mathbf{k}})\cdot\hat{\mathbf{k}}(b\,dy)\\&=&cb\int_0^ay^2dy=\frac{1}{3}a^3bc.\end{eqnarray*}$

Significance

For a non-constant electric field, the integral method is required.

CHECK YOUR UNDERSTANDING 6.2

If the electric field in Example 2.1.4 is $\vec{\mathbf{E}}=mx\hat{\mathbf{k}}$ what is the flux through the rectangular area?

12

2.2 Explaining Gauss’s Law

LEARNING OBJECTIVES

By the end of this section, you will be able to:

State Gauss’s law
Explain the conditions under which Gauss’s law may be used
Apply Gauss’s law in appropriate systems

We can now determine the electric flux through an arbitrary closed surface due to an arbitrary charge distribution. We found that if a closed surface does not have any charge inside where an electric field line can terminate, then any electric field line entering the surface at one point must necessarily exit at some other point of the surface. Therefore, if a closed surface does not have any charges inside the enclosed volume, then the electric flux through the surface is zero. Now, what happens to the electric flux if there are some charges inside the enclosed volume? Gauss’s law gives a quantitative answer to this question.

To get a feel for what to expect, let’s calculate the electric flux through a spherical surface around a positive point charge q, since we already know the electric field in such a situation. Recall that when we place the point charge at the origin of a coordinate system, the electric field at a point P that is at a distance r from the charge at the origin is given by

$\vec{\mathbf{E}}_P=\frac{1}{4\pi\epsilon_0}\frac{1}{r^2}\hat{\mathbf{r}},$

where $\hat{\mathbf{r}}$ is the radial vector from the charge at the origin to the point $P$ . We can use this electric field to find the flux through the spherical surface of radius $r$ , as shown in Figure 2.2.1.

(Figure 2.2.1) $\begin{gather*}.\end{gather*}$

A sphere labeled S with radius R is shown. At its center, is a small circle with a plus sign, labeled q. A small patch on the sphere is labeled dA. Two arrows point outward from here, perpendicular to the surface of the sphere. The smaller arrow is labeled n hat equal to r hat. The longer arrow is labeled vector E.

Figure 2.2.1 A closed spherical surface surrounding a point charge $q$ .

Then we apply $\Phi=\int_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA$ to this system and substitute known values. On the sphere, $\hat{\mathbf{n}}=\hat{\mathbf{r}}$ and $r=R$ , so for an infinitesimal area $dA$ ,

$d\Phi=\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}dA=\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}dA.$

We now find the net flux by integrating this flux over the surface of the sphere:

$\Phi=\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}\oint_SdA=\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}(4\pi R^2)=\frac{q}{\epsilon_0}.$

where the total surface area of the spherical surface is $4\piR^2$ This gives the flux through the closed spherical surface at radius $r$ as

(2.2.1) $\begin{equation*}\Phi=\frac{q}{\epsilon_0}.\end{equation*}$

A remarkable fact about this equation is that the flux is independent of the size of the spherical surface. This can be directly attributed to the fact that the electric field of a point charge decreases as $1/r^2$ with distance, which just cancels the $r^2$ rate of increase of the surface area.

Electric Field Lines Picture

An alternative way to see why the flux through a closed spherical surface is independent of the radius of the surface is to look at the electric field lines. Note that every field line from $q$ that pierces the surface at radius $R_1$ also pierces the surface at $R_2$ (Figure 2.2.2).

(Figure 2.2.2) $\begin{gather*}.\end{gather*}$

Figure shows three concentric circles. The smallest one at the center is labeled q, the middle one has radius R1 and the largest one has radius R2. Eight arrows radiate outward from the center in all eight directions.

Figure 2.2.2 Flux through spherical surfaces of radii $R_1$ and $R_2$ enclosing a charge $q$ are equal, independent of the size of the surface, since all $E$ -field lines that pierce one surface from the inside to outside direction also pierce the other surface in the same direction.

Therefore, the net number of electric field lines passing through the two surfaces from the inside to outside direction is equal. This net number of electric field lines, which is obtained by subtracting the number of lines in the direction from outside to inside from the number of lines in the direction from inside to outside gives a visual measure of the electric flux through the surfaces.

You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. A typical field line enters the surface at $dA_1$ and leaves at $dA_2$ Every line that enters the surface must also leave that surface. Hence the net “flow” of the field lines into or out of the surface is zero (Figure 2.2.3(a)). The same thing happens if charges of equal and opposite sign are included inside the closed surface, so that the total charge included is zero (part (b)). A surface that includes the same amount of charge has the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surface encloses the same amount of charge (part (c)).

(Figure 2.2.3) $\begin{gather*}.\end{gather*}$

Figure a shows an irregular 3 dimensional shape labeled S. A small circle with a plus sign, labeled q is outside it. Three arrows labeled vector E originate from q and pass through S. The patches where the arrows pierce the surface of S are highlighted. The patch where one arrow enters the shape is labeled dA1 and the patch where the arrow emerges from the shape is labeled dA2. Figure b shows an oval with two small circles inside it. These are labeled plus and minus. Three arrow from outside the oval point to the circle labeled minus. Three arrows point from plus to minus. Three arrows point from plus to outside the oval. Figure c has an irregular shape labeled S2. Within it is a circle named S1. At its center is a small circle labeled plus. Six arrows radiate outward from here in different directions.

Figure 2.2.3 Understanding the flux in terms of field lines. (a) The electric flux through a closed surface due to a charge outside that surface is zero. (b) Charges are enclosed, but because the net charge included is zero, the net flux through the closed surface is also zero. (c) The shape and size of the surfaces that enclose a charge does not matter because all surfaces enclosing the same charge have the same flux.

Statement of Gauss’s Law

Gauss’s law generalizes this result to the case of any number of charges and any location of the charges in the space inside the closed surface. According to Gauss’s law, the flux of the electric field $\vec{\mathbf{E}}$ through any closed surface, also called a Gaussian surface, is equal to the net charge enclosed ( $q_{\mathrm{enc}}$ ) divided by the permittivity of free space ( $\epsilon_0$ ):

$\Phi_{\mathrm{Closed~Surface}}=\frac{q_{\mathrm{enc}}}{\epsilon_0}.$

This equation holds for charges of either sign, because we define the area vector of a closed surface to point outward. If the enclosed charge is negative (see Figure 2.2.4(b)), then the flux through either $S$ or $S'$ is negative.

(Figure 2.2.4) $\begin{gather*}.\end{gather*}$

Figure a has an irregular shape labeled S. Within it is a circle labeled S prime. At its center is a small circle labeled plus. Six arrows radiate outward from here in different directions. Figure b has the same irregular shape S and circle S prime. At its center is a small circle labeled minus. Six arrows from different directions radiate inward to minus.

Figure 2.2.4 The electric flux through any closed surface surrounding a point charge $q$ is given by Gauss’s law. (a) Enclosed charge is positive. (b) Enclosed charge is negative.

The Gaussian surface does not need to correspond to a real, physical object; indeed, it rarely will. It is a mathematical construct that may be of any shape, provided that it is closed. However, since our goal is to integrate the flux over it, we tend to choose shapes that are highly symmetrical.

If the charges are discrete point charges, then we just add them. If the charge is described by a continuous distribution, then we need to integrate appropriately to find the total charge that resides inside the enclosed volume. For example, the flux through the Gaussian surface $S$ of Figure 2.2.5 is $\Phi=(q_1+q_2+q_5)/\epsilon_0$ . Note that $q_{\mathrm{enc}}$ is simply the sum of the point charges. If the charge distribution were continuous, we would need to integrate appropriately to compute the total charge within the Gaussian surface.

(Figure 2.2.5) $\begin{gather*}.\end{gather*}$

Figure shows an irregular shape labeled S. Within it are charges labeled positive q1 and negative q2 and q5. Outside S are charges labeled positive q3, q4, q6 and q N minus 1 and negative q7 and q N.

Figure 2.2.5 The flux through the Gaussian surface shown, due to the charge distribution, is $\Phi=(q_1+q_2+q_5)/\epsilon_0$ .

Recall that the principle of superposition holds for the electric field. Therefore, the total electric field at any point, including those on the chosen Gaussian surface, is the sum of all the electric fields present at this point. This allows us to write Gauss’s law in terms of the total electric field.

GAUSS’S LAW

The flux $\Phi$ of the electric field $\vec{\mathbf{E}}$ through any closed surface $S$ (a Gaussian surface) is equal to the net charge enclosed ( $q_{\mathrm{enc}}$ ) divided by the permittivity of free space ( $\epsilon_0$ ):

(2.2.2) $\begin{equation*}\Phi=\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\frac{q_{\mathrm{enc}}}{\epsilon_0}.\end{equation*}$

To use Gauss’s law effectively, you must have a clear understanding of what each term in the equation represents. The field $\vec{\mathbf{E}}$ is the total electric field at every point on the Gaussian surface. This total field includes contributions from charges both inside and outside the Gaussian surface. However, $q_{\mathrm{enc}}$ is just the charge inside the Gaussian surface. Finally, the Gaussian surface is any closed surface in space. That surface can coincide with the actual surface of a conductor, or it can be an imaginary geometric surface. The only requirement imposed on a Gaussian surface is that it be closed (Figure 2.2.6).

(Figure 2.2.6) $\begin{gather*}.\end{gather*}$

Figure shows a bottle that looks like an upside down flask whose neck is elongated, bent upward, twisted, taken inside the bottle and joined with its base, thus having only one surface.

Figure 2.2.6 A Klein bottle partially filled with a liquid. Could the Klein bottle be used as a Gaussian surface?

EXAMPLE 2.2.1

Electric Flux through Gaussian Surfaces

Calculate the electric flux through each Gaussian surface shown in Figure 2.2.7.

(Figure 2.2.7) $\begin{gather*}.\end{gather*}$

Figures a through d show irregular shapes and figure e shows a cube. Figure a has a charge inside the shape labeled plus 2.0 mu C. Figure b has a charge inside the shape labeled minus 2.0 mu C. Figure c has a charge inside the shape labeled plus 2.0 mu C and two charges outside labeled plus 4 mu C and minus 2.0 mu C. Figure d has three charges inside the shape labeled minus 1.0 mu C, minus 4.0 mu C and plus 6.0 mu C and two charges outside the shape labeled minus 5.0 mu C and plus 4.0 mu C. Figure e has three charges inside labeled plus 4.0 mu C, plus 6.0 mu C and minus 10.0 mu C and two charges outside the cube labeled plus 5.0 mu C and 3.0 mu C.

Figure 2.2.7 Various Gaussian surfaces and charges.

Strategy

From Gauss’s law, the flux through each surface is given by $q_{\mathrm{enc}}/\epsilon_0$ , where $q_{\mathrm{enc}}$ is the charge enclosed by that surface.

Solution

For the surfaces and charges shown, we find

$\Phi=\frac{2.0~\mu\mathrm{C}}{\epsilon_0}=2.3\times10^5~\mathrm{N}\cdot\mathrm^{m}^2/\mathrm{C}$ .
$\Phi=\frac{-2.0~\mu\mathrm{C}}{\epsilon_0}=-2.3\times10^5~\mathrm{N}\cdot\mathrm^{m}^2/\mathrm{C}$ .
$\Phi=\frac{2.0~\mu\mathrm{C}}{\epsilon_0}=2.3\times10^5~\mathrm{N}\cdot\mathrm^{m}^2/\mathrm{C}$ .
$\Phi=\frac{-4.0~\mu\mathrm{C}+6.0~\mu\mathrm{C}-1.0~\mu\mathrm{C}}{\epsilon_0}=1.1\times10^5~\mathrm{N}\cdot\mathrm^{m}^2/\mathrm{C}$ .
$\Phi=\frac{4.0~\mu\mathrm{C}+6.0~\mu\mathrm{C}-10.0~\mu\mathrm{C}}{\epsilon_0}=0$ .

Significance

In the special case of a closed surface, the flux calculations become a sum of charges. In the next section, this will allow us to work with more complex systems.

CHECK YOUR UNDERSTANDING 2.3

Calculate the electric flux through the closed cubical surface for each charge distribution shown in Figure 2.2.8.

(Figure 2.2.8) $\begin{gather*}.\end{gather*}$

Figures a through d show a cuboid with one corner at the origin of the coordinate axes. In figure a, there is a charge plus 3.0 mu C on the surface parallel to the yz plane. In figure b, there is a charge minus 3.0 mu C on the surface parallel to the yz plane. In figure c, there is a charge plus 3.0 mu C on the surface parallel to the yz plane, a charge minus 3.0 mu C on the y axis outside the shape and a charge plus 6.0 mu C outside the shape. In figure d, there is a charge minus 3.0 mu C on the y axis outside the shape and charges plus 3.0 mu C and plus 6.0 mu C outside the shape.

Figure 2.2.8 A cubical Gaussian surface with various charge distributions.

INTERACTIVE

Use this simulation to adjust the magnitude of the charge and the radius of the Gaussian surface around it. See how this affects the total flux and the magnitude of the electric field at the Gaussian surface.

13

2.3 Applying Gauss’s Law

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Explain what spherical, cylindrical, and planar symmetry are
Recognize whether or not a given system possesses one of these symmetries
Apply Gauss’s law to determine the electric field of a system with one of these symmetries

Gauss’s law is very helpful in determining expressions for the electric field, even though the law is not directly about the electric field; it is about the electric flux. It turns out that in situations that have certain symmetries (spherical, cylindrical, or planar) in the charge distribution, we can deduce the electric field based on knowledge of the electric flux. In these systems, we can find a Gaussian surface $S$ over which the electric field has constant magnitude. Furthermore, if $\vec{\mathbf{E}}$ is parallel to $\hat{\mathbf{n}}$ everywhere on the surface, then $\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}=E$ . (If $\vec{\mathbf{E}}$ and $\hat{\mathbf{n}}$ are antiparallel everywhere on the surface, then $\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}=-E$ .) Gauss’s law then simplifies to

(2.3.1) $\begin{equation*}\Phi=\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=E\oint_SdA=EA=\frac{q_{\mathrm{enc}}}{\epsilon_0},\end{equation*}$

where $A$ is the area of the surface. Note that these symmetries lead to the transformation of the flux integral into a product of the magnitude of the electric field and an appropriate area. When you use this flux in the expression for Gauss’s law, you obtain an algebraic equation that you can solve for the magnitude of the electric field, which looks like

$E\sim\frac{q_{\mathrm{enc}}}{\epsilon_0~\mathrm{area}}.$

The direction of the electric field at the field point $P$ is obtained from the symmetry of the charge distribution and the type of charge in the distribution. Therefore, Gauss’s law can be used to determine $\vec{\mathbf{E}}$ . Here is a summary of the steps we will follow:

Problem-Solving Strategy: Gauss’s Law

Identify the spatial symmetry of the charge distribution. This is an important first step that allows us to choose the appropriate Gaussian surface. As examples, an isolated point charge has spherical symmetry, and an infinite line of charge has cylindrical symmetry.
Choose a Gaussian surface with the same symmetry as the charge distribution and identify its consequences. With this choice, $\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}$ is easily determined over the Gaussian surface.
Evaluate the integral \oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA over the Gaussian surface, that is, calculate the flux through the surface. The symmetry of the Gaussian surface allows us to factor $\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}$ outside the integral.
Determine the amount of charge enclosed by the Gaussian surface. This is an evaluation of the right-hand side of the equation representing Gauss’s law. It is often necessary to perform an integration to obtain the net enclosed charge.
Evaluate the electric field of the charge distribution. The field may now be found using the results of steps 3 and 4.

Basically, there are only three types of symmetry that allow Gauss’s law to be used to deduce the electric field. They are

A charge distribution with spherical symmetry
A charge distribution with cylindrical symmetry
A charge distribution with planar symmetry

To exploit the symmetry, we perform the calculations in appropriate coordinate systems and use the right kind of Gaussian surface for that symmetry, applying the remaining four steps.

Charge Distribution with Spherical Symmetry

A charge distribution has spherical symmetry if the density of charge depends only on the distance from a point in space and not on the direction. In other words, if you rotate the system, it doesn’t look different. For instance, if a sphere of radius $R$ is uniformly charged with charge density $\rho_0$ then the distribution has spherical symmetry (Figure 2.3.1(a)). On the other hand, if a sphere of radius $R$ is charged so that the top half of the sphere has uniform charge density $\rho_1$ and the bottom half has a uniform charge density $\rho_2\neq\rho_1$ , then the sphere does not have spherical symmetry because the charge density depends on the direction (Figure 2.3.1(b)). Thus, it is not the shape of the object but rather the shape of the charge distribution that determines whether or not a system has spherical symmetry.

Figure 2.3.1(c) shows a sphere with four different shells, each with its own uniform charge density. Although this is a situation where charge density in the full sphere is not uniform, the charge density function depends only on the distance from the centre and not on the direction. Therefore, this charge distribution does have spherical symmetry.

(Figure 2.3.1) $\begin{gather*}.\end{gather*}$

Figure a shows a uniformly colored sphere labeled rho 0. The figure is labeled spherically symmetric. Figure b shows a sphere whose top and bottom halves are differently colored. The top hemisphere is labeled rho 1 and the bottom one is labeled rho 2. The figure is labeled not spherically symmetric. Figure c shows a sphere, sectioned to show many concentric spheres of different colors within it. The figure is labeled spherically symmetric.

Figure 2.3.1 Illustrations of spherically symmetrical and nonsymmetrical systems. Different shadings indicate different charge densities. Charges on spherically shaped objects do not necessarily mean the charges are distributed with spherical symmetry. The spherical symmetry occurs only when the charge density does not depend on the direction. In (a), charges are distributed uniformly in a sphere. In (b), the upper half of the sphere has a different charge density from the lower half; therefore, (b) does not have spherical symmetry. In (c), the charges are in spherical shells of different charge densities, which means that charge density is only a function of the radial distance from the centre; therefore, the system has spherical symmetry.

One good way to determine whether or not your problem has spherical symmetry is to look at the charge density function in spherical coordinates, $\rho(r,\theta,\phi)$ . If the charge density is only a function of $r$ , that is $\rho=\rho(r)$ , then you have spherical symmetry. If the density depends on $\theta$ or $\phi$ you could change it by rotation; hence, you would not have spherical symmetry.

Consequences of symmetry

In all spherically symmetrical cases, the electric field at any point must be radially directed, because the charge and, hence, the field must be invariant under rotation. Therefore, using spherical coordinates with their origins at the centre of the spherical charge distribution, we can write down the expected form of the electric field at a point $P$ located at a distance $r$ from the centre:

(2.3.2) $\begin{equation*}\mathrm{Spherical~symmetry}:~\vec{\mathbf{E}}_P=E_P(r)\hat{\mathbf{r}},\end{equation*}$

where $\hat{\mathbf{r}}$ is the unit vector pointed in the direction from the origin to the field point $P$ . The radial component $E_P$ of the electric field can be positive or negative. When $E_P>0$ , the electric field at $P$ points away from the origin, and when $E_P<0$ , the electric field at $P$ points toward the origin.

Gaussian surface and flux calculations

We can now use this form of the electric field to obtain the flux of the electric field through the Gaussian surface. For spherical symmetry, the Gaussian surface is a closed spherical surface that has the same centre as the centre of the charge distribution. Thus, the direction of the area vector of an area element on the Gaussian surface at any point is parallel to the direction of the electric field at that point, since they are both radially directed outward (Figure 2.3.2).

(Figure 2.3.2) $\begin{gather*}.\end{gather*}$

Figure shows a circle labeled charges with center O and radius R. A larger concentric circle shown with a dotted line is labeled Gaussian surface. An arrow labeled r marks the radius of the outer circle. An arrow labeled r hat is shown along r. A small patch where r touches the Gaussian surface is highlighted and labeled P. From here, another arrow points outward in the same direction as r. This is labeled vector E subscript P. Another arrow originates from the tip of vector E subscript P and points outward in the same direction. It is labeled delta vector A.

Figure 2.3.2 The electric field at any point of the spherical Gaussian surface for a spherically symmetrical charge distribution is parallel to the area element vector at that point, giving flux as the product of the magnitude of electric field and the value of the area. Note that the radius $R$ of the charge distribution and the radius $r$ of the Gaussian surface are different quantities.

The magnitude of the electric field $\vec{\mathbf{E}}$ must be the same everywhere on a spherical Gaussian surface concentric with the distribution. For a spherical surface of radius $r$ ,

$\Phi=\oint_S\vec{\mathbf{E}}_P\cdot\hat{\mathbf{n}}dA=E_P\oint_SdA=E_P4\pi r^2.$

Using Gauss’s law

According to Gauss’s law, the flux through a closed surface is equal to the total charge enclosed within the closed surface divided by the permittivity of vacuum $\epsilon_0$ . Let $q_{\mathrm{enc}}$ be the total charge enclosed inside the distance $r$ from the origin, which is the space inside the Gaussian spherical surface of radius $r$ . This gives the following relation for Gauss’s law:

$4\pi r^2E=\frac{q_{\mathrm{enc}}}{\epsilon_0}.$

Hence, the electric field at point $P$ that is a distance $r$ from the centre of a spherically symmetrical charge distribution has the following magnitude and direction:

(2.3.3) $\begin{equation*}\mathrm{Magnitude}:~E(r)=\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{enc}}}{r^2}\end{equation*}$

Direction: radial from $O$ to $P$ or from $P$ to $O$ .

The direction of the field at point $P$ depends on whether the charge in the sphere is positive or negative. For a net positive charge enclosed within the Gaussian surface, the direction is from $O$ to $P$ , and for a net negative charge, the direction is from $P$ to $O$ . This is all we need for a point charge, and you will notice that the result above is identical to that for a point charge. However, Gauss’s law becomes truly useful in cases where the charge occupies a finite volume.

Computing enclosed charge

The more interesting case is when a spherical charge distribution occupies a volume, and asking what the electric field inside the charge distribution is thus becomes relevant. In this case, the charge enclosed depends on the distance $r$ of the field point relative to the radius of the charge distribution $R$ , such as that shown in Figure 2.3.3.

(Figure 2.3.3) $\begin{gather*}.\end{gather*}$

Figure a shows a dotted circle S with center O and radius r, and a larger concentric circle with radius R. A small arrow points outward from S. This is labeled vector E subscript in. S is labeled Gaussian surface for vector E subscript in. Figure b shows a dotted circle S with center O and radius r, and a smaller concentric circle with radius R. A small arrow points outward from S. This is labeled vector E subscript out. S is labeled Gaussian surface for E vector subscript out.

Figure 2.3.3 A spherically symmetrical charge distribution and the Gaussian surface used for finding the field (a) inside and (b) outside the distribution.

If point $P$ is located outside the charge distribution—that is, if $r\geq R$ —then the Gaussian surface containing $P$ encloses all charges in the sphere. In this case, $q_{\mathrm{enc}}$ equals the total charge in the sphere. On the other hand, if point $P$ is within the spherical charge distribution, that is, if $r<R$ , then the Gaussian surface encloses a smaller sphere than the sphere of charge distribution. In this case, $q_{\mathrm{enc}}$ is less than the total charge present in the sphere. Referring to Figure 2.3.3, we can write $q_{\mathrm{enc}}$ as

$q_{\mathrm{enc}}=\left\{\begin{array}{l}q_{\mathrm{tot}}~(\mathrm{total~charge})~\mathrm{if}~r\geq R\\q_{\mathrm{within}~r<R}~(\mathrm{only~charge~within}~r<R)~\mathrm{if}~r<R\end{array}\right..$

The field at a point outside the charge distribution is also called $\vec{\mathbf{E}}_{\mathrm{out}$ , and the field at a point inside the charge distribution is called $\vec{\mathbf{E}}_{\mathrm{in}$ . Focusing on the two types of field points, either inside or outside the charge distribution, we can now write the magnitude of the electric field as

(2.3.4) $\begin{equation*}P~\mathrm{outside~sphere}:~E_{\mathrm{out}}=\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{tot}}}{r^2}\end{equation*}$

(2.3.5) $\begin{equation*}P~\mathrm{inside~sphere}:~E_{\mathrm{in}}=\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{within}~r<R}}{r^2}.\end{equation*}$

Note that the electric field outside a spherically symmetrical charge distribution is identical to that of a point charge at the centre that has a charge equal to the total charge of the spherical charge distribution. This is remarkable since the charges are not located at the centre only. We now work out specific examples of spherical charge distributions, starting with the case of a uniformly charged sphere.

EXAMPLE 2.3.1

Uniformly Charged Sphere

A sphere of radius $R$ , such as that shown in Figure 2.3.3, has a uniform volume charge density $\rho_0$ . Find the electric field at a point outside the sphere and at a point inside the sphere.

Strategy

Apply the Gauss’s law problem-solving strategy, where we have already worked out the flux calculation.

Solution

The charge enclosed by the Gaussian surface is given by

$q_{\mathrm{enc}}=\int\rho_{0}dV=\int_0^r\rho_04\pi r'^2dr'=\rho_0\left(\frac{4}{3}\pi r^3\right).$

The answer for electric field amplitude can then be written down immediately for a point outside the sphere, labeled $E_{\mathrm{out}}$ , and a point inside the sphere, labeled $E_{\mathrm{in}}$ .

$\begin{eqnarray*}E_{\mathrm{out}}&=&\frac{1}{4\pi\epsilon_0}\frac{q_{\mathrm{tot}}}{r^2},\ q_{\mathrm{tot}}=\frac{4}{3}\pi R^3\rho_0,\\E_{\mathrm{in}}&=&\frac{q_{\mathrm{enc}}}{4\pi\epsilon_0r^2}=\frac{\rho_0r}{3\epsilon_0},\ q_{\mathrm{enc}}=\frac{4}{3}\pi r^3\rho_0.\end{eqnarray*}$

It is interesting to note that the magnitude of the electric field increases inside the material as you go out, since the amount of charge enclosed by the Gaussian surface increases with the volume. Specifically, the charge enclosed grows $\propto r^3$ , whereas the field from each infinitesimal element of charge drops off $\propto 1/r^2$ with the net result that the electric field within the distribution increases in strength linearly with the radius. The magnitude of the electric field outside the sphere decreases as you go away from the charges, because the included charge remains the same but the distance increases. Figure 2.3.4 displays the variation of the magnitude of the electric field with distance from the centre of a uniformly charged sphere.

(Figure 2.3.4) $\begin{gather*}.\end{gather*}$

Figure shows a graph of E versus r. The curve rises in a straight line labeled E proportional to r, peaks and falls in a curved line labeled E proportional to 1 by r squared. The peak has an x value of R and a y value of E subscript R.

Figure 2.3.4 Electric field of a uniformly charged, non-conducting sphere increases inside the sphere to a maximum at the surface and then decreases as $1/r^2$ . Here, $E_R=\frac{\rho_0R}{3\epsilon_0}$ . The electric field is due to a spherical charge distribution of uniform charge density and total charge $Q$ as a function of distance from the centre of the distribution.

The direction of the electric field at any point $P$ is radially outward from the origin if $\rho_0$ is positive, and inward (i.e., toward the centre) if $\rho_0$ is negative. The electric field at some representative space points are displayed in Figure 2.3.5 whose radial coordinates $r$ are $r=R/2$ , $r=R$ , and $r=2R$ .

(Figure 2.3.5) $\begin{gather*}.\end{gather*}$

Figure shows three concentric circles. The smallest one is dotted and labeled r equal to R by 2. The middle one is labeled r equal to R and the largest one, also dotted, is labeled r equal to 2R. Arrows labeled vector E originate from each circle and point outward, perpendicular to the circle. The ones on the outer circle are smallest and the ones on the middle circle are the longest.

Figure 2.3.5 Electric field vectors inside and outside a uniformly charged sphere.

Significance

Notice that $E_{\mathrm{out}}$ has the same form as the equation of the electric field of an isolated point charge. In determining the electric field of a uniform spherical charge distribution, we can therefore assume that all of the charge inside the appropriate spherical Gaussian surface is located at the centre of the distribution.

EXAMPLE 2.3.2

Non-Uniformly Charged Sphere

A non-conducting sphere of radius $R$ has a non-uniform charge density that varies with the distance from its centre as given by

$\rho(r)=ar^n~(r\leq R;~n\geq0),$

where $a$ is a constant. We require $n\geq0$ so that the charge density is not undefined at $r=0$ . Find the electric field at a point outside the sphere and at a point inside the sphere.

Strategy

Apply the Gauss’s law strategy given above, where we work out the enclosed charge integrals separately for cases inside and outside the sphere.

Solution

Since the given charge density function has only a radial dependence and no dependence on direction, we have a spherically symmetrical situation. Therefore, the magnitude of the electric field at any point is given above and the direction is radial. We just need to find the enclosed charge $q_{\mathrm{enc}}$ , which depends on the location of the field point.

A note about symbols: We use $r'$ for locating charges in the charge distribution and $r$ for locating the field point(s) at the Gaussian surface(s). The letter $R$ is used for the radius of the charge distribution.

As charge density is not constant here, we need to integrate the charge density function over the volume enclosed by the Gaussian surface. Therefore, we set up the problem for charges in one spherical shell, say between $r'$ and $r'+dr'$ , as shown in Figure 2.3.6 . The volume of charges in the shell of infinitesimal width is equal to the product of the area of surface $4\pi r'^2$ and the thickness $dr'$ . Multiplying the volume with the density at this location, which is $ar'^n$ , gives the charge in the shell:

$dq=ar'^n4\pi r'^2dr'.$

(Figure 2.3.6) $\begin{gather*}.\end{gather*}$

Figure shows four concentric circles. Starting from the smallest, their radii are labeled: r prime, r prime plus d r prime, R and r. The outermost circle is dotted and labeled Gaussian surface.

Figure 2.3.6 Spherical symmetry with non-uniform charge distribution. In this type of problem, we need four radii: $R$ is the radius of the charge distribution, $r$ is the radius of the Gaussian surface, $r'$ is the inner radius of the spherical shell, and $r'+dr'$ is the outer radius of the spherical shell. The spherical shell is used to calculate the charge enclosed within the Gaussian surface. The range for $r'$ is from $0$ to $r$ for the field at a point inside the charge distribution and from $0$ to $R$ for the field at a point outside the charge distribution. If $r>R$ , then the Gaussian surface encloses more volume than the charge distribution, but the additional volume does not contribute to $q_{\mathrm{enc}}$ .

(a) Field at a point outside the charge distribution. In this case, the Gaussian surface, which contains the field point $P$ , has a radius $r$ that is greater than the radius $R$ of the charge distribution, $r>R$ . Therefore, all charges of the charge distribution are enclosed within the Gaussian surface. Note that the space between $r'=R$ and $r'=r$ is empty of charges and therefore does not contribute to the integral over the volume enclosed by the Gaussian surface:

$q_{\mathrm{enc}}=\int dq=\int_0^Rar'^n4\pir'^2dr'=\frac{4\pi a}{n+3}R^{n+3}.$

This is used in the general result for $\vec{\mathbf{E}}_{\mathrm{out}}$ above to obtain the electric field at a point outside the charge distribution as

$\vec{\mathbf{E}}_{\mathrm{out}}=\left[\frac{aR^{n+3}}{\epsilon_0(n+3)}\right]\frac{1}{r^2}\hat{\mathbf{r}},$

where $\hat{\mathbf{r}}$ is a unit vector in the direction from the origin to the field point at the Gaussian surface.

(b) Field at a point inside the charge distribution. The Gaussian surface is now buried inside the charge distribution, with $r<R$ . Therefore, only those charges in the distribution that are within a distance $r$ of the centre of the spherical charge distribution count in $r_{\mathrm{enc}}$ :

$q_{\mathrm{enc}}=\int_0^rar'^n4\pir'^2dr'=\frac{4\pi a}{n+3}r^{n+3}.$

Now, using the general result above for $\vec{\mathbf{E}}_{\mathrm{in}}$ we find the electric field at a point that is a distance $r$ from the centre and lies within the charge distribution as

$\vec{\mathbf{E}}_{\mathrm{in}}=\left[\frac{a}{\epsilon_0(n+3)}\right]r^{n+1}\hat{\mathbf{r}},$

where the direction information is included by using the unit radial vector.

CHECK YOUR UNDERSTANDING 2.4

Check that the electric fields for the sphere reduce to the correct values for a point charge.

Charge Distribution with Cylindrical Symmetry

A charge distribution has cylindrical symmetry if the charge density depends only upon the distance $r$ from the axis of a cylinder and must not vary along the axis or with direction about the axis. In other words, if your system varies if you rotate it around the axis, or shift it along the axis, you do not have cylindrical symmetry.

Figure 2.3.7 shows four situations in which charges are distributed in a cylinder. A uniform charge density $\rho_0$ . in an infinite straight wire has a cylindrical symmetry, and so does an infinitely long cylinder with constant charge density $\rho_0$ . An infinitely long cylinder that has different charge densities along its length, such as a charge density $\rho_1$ for $z>0$ and $\rho_2\neq\rho_1$ for $z<0$ , does not have a usable cylindrical symmetry for this course. Neither does a cylinder in which charge density varies with the direction, such as a charge density $\rho_1$ for $0\leq\theta<\pi$ and $\rho_2\neq\rho_1$ for $\pi\leq\theta<2\pi$ . A system with concentric cylindrical shells, each with uniform charge densities, albeit different in different shells, as in Figure 2.3.7(d), does have cylindrical symmetry if they are infinitely long. The infinite length requirement is due to the charge density changing along the axis of a finite cylinder. In real systems, we don’t have infinite cylinders; however, if the cylindrical object is considerably longer than the radius from it that we are interested in, then the approximation of an infinite cylinder becomes useful.

(Figure 2.3.7) $\begin{gather*}.\end{gather*}$

Figures a through d show a cylinder. In figure a, labeled cylindrically symmetrical, the cylinder is uniformly colored and labeled rho zero. In figure b, labeled not cylindrically symmetrical, the top and bottom halves of the cylinder are different in color. The top is labeled rho 1 and the bottom is labeled rho 2. In figure c, labeled not cylindrically symmetrical, the left and right halves of the cylinder are different in color. The left is labeled rho 1 and the right is labeled rho 2. In figure d, many concentric sections are seen within the cylinder. The figure is labeled cylindrically symmetrical.

Figure 2.3.7 To determine whether a given charge distribution has cylindrical symmetry, look at the cross-section of an “infinitely long” cylinder. If the charge density does not depend on the polar angle of the cross-section or along the axis, then you have cylindrical symmetry. (a) Charge density is constant in the cylinder; (b) upper half of the cylinder has a different charge density from the lower half; (c) left half of the cylinder has a different charge density from the right half; (d) charges are constant in different cylindrical rings, but the density does not depend on the polar angle. Cases (a) and (d) have cylindrical symmetry, whereas (b) and (c) do not.

Consequences of symmetry

In all cylindrically symmetrical cases, the electric field $\vec{\mathbf{E}}_P$ at any point $P$ must also display cylindrical symmetry.

Cylindrical symmetry: $\vec{\mathbf{E}}_P=E_P(r)\hat{\mathbf{r}}$ ,

where $r$ is the distance from the axis and $\hat{\mathbf{r}}$ is a unit vector directed perpendicularly away from the axis (Figure 2.3.8).

(Figure 2.3.8) $\begin{gather*}.\end{gather*}$

A cylinder is shown with a dotted line. A circular portion within the cylinder, at its center is highlighted. The radius of the circle and that of the cylinder is labeled r. The point where r touches the cylinder is labeled P. An arrow labeled r hat originates from P and points outward in the same line as r.

Figure 2.3.8 The electric field in a cylindrically symmetrical situation depends only on the distance from the axis. The direction of the electric field is pointed away from the axis for positive charges and toward the axis for negative charges.

Gaussian surface and flux calculation

To make use of the direction and functional dependence of the electric field, we choose a closed Gaussian surface in the shape of a cylinder with the same axis as the axis of the charge distribution. The flux through this surface of radius $s$ and height $L$ is easy to compute if we divide our task into two parts: (a) a flux through the flat ends and (b) a flux through the curved surface (Figure 2.3.9 ).

(Figure 2.3.9) $\begin{gather*}.\end{gather*}$

Figure shows a cylinder of length L. A line perpendicular to the axis connects the axis to point P on the surface of the cylinder. An arrow labeled delta vector A points outward from P in the same direction as the line. Another arrow labeled vector E subscript P originates from the tip of the first arrow and points in the same direction. A third arrow labeled delta vector A points outward from the top surface of the cylinder, perpendicular to it. An arrow labeled vector E originates from the base of the third arrow and is perpendicular to it.

Figure 2.3.9 The Gaussian surface in the case of cylindrical symmetry. The electric field at a patch is either parallel or perpendicular to the normal to the patch of the Gaussian surface.

The electric field is perpendicular to the cylindrical side and parallel to the planar end caps of the surface. The flux through the cylindrical part is

$\int_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=E\int_SdA=E(2\pi rL),$

whereas the flux through the end caps is zero because $\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}=0$ there. Thus, the flux is

$\int_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=E(2\pi rL)+0+0=2\pi rLE.$

Using Gauss’s law

According to Gauss’s law, the flux must equal the amount of charge within the volume enclosed by this surface, divided by the permittivity of free space. When you do the calculation for a cylinder of length $L$ , you find that $q_{\mathrm{enc}}$ of Gauss’s law is directly proportional to $L$ . Let us write it as charge per unit length ( $\lambda_{\mathrm{enc}}$ ) times length $L$ :

$q_{\mathrm{enc}}=\lambda_{\mathrm{enc}}L.$

Hence, Gauss’s law for any cylindrically symmetrical charge distribution yields the following magnitude of the electric field a distance $s$ away from the axis:

$\mathrm{Magnitude}:~E(r)=\frac{\lambda_{\mathrm{enc}}}{2\pi\epsilon_0}\frac{1}{r}.$

The charge per unit length $\lambda_{\mathrm{enc}}$ depends on whether the field point is inside or outside the cylinder of charge distribution, just as we have seen for the spherical distribution.

Computing enclosed charge

Let $R$ be the radius of the cylinder within which charges are distributed in a cylindrically symmetrical way. Let the field point $P$ be at a distance s from the axis. (The side of the Gaussian surface includes the field point $P$ .) When $r>R$ (that is, when $P$ is outside the charge distribution), the Gaussian surface includes all the charge in the cylinder of radius $R$ and length $L$ . When $r<R$ ( $P$ is located inside the charge distribution), then only the charge within a cylinder of radius $s$ and length $L$ is enclosed by the Gaussian surface:

$\lambda_{\mathrm{enc}}L=\left\{\begin{array}{l}(\mathrm{total~charge})~\mathrm{if}~r\geq R\\(\mathrm{only~charge~within}~r<R)~\mathrm{if}~r<R\end{array}\right..$

EXAMPLE 2.3.3

Uniformly Charged Cylindrical Shell

A very long non-conducting cylindrical shell of radius $R$ has a uniform surface charge density $\sigma_0$ . Find the electric field (a) at a point outside the shell and (b) at a point inside the shell.

Strategy

Apply the Gauss’s law strategy given earlier, where we treat the cases inside and outside the shell separately.

Solution

(a) Electric field at a point outside the shell. For a point outside the cylindrical shell, the Gaussian surface is the surface of a cylinder of radius $r>R$ and length $L$ , as shown in Figure 2.3.10. The charge enclosed by the Gaussian cylinder is equal to the charge on the cylindrical shell of length $L$ . Therefore, $\lambda_{\mathrm{enc}}$ is given by

$\lambda_{\mathrm{enc}}=\frac{\sigma_02\pi RL}{L}=2\pi R\sigma_0.$

(Figure 2.3.10) $\begin{gather*}.\end{gather*}$

Two cylinders sharing the same axis are shown. The outer one has length L, which is smaller than the inner cylinder’s length. A line perpendicular to the axis connects the axis to point P on the surface of the outer cylinder. An arrow labeled r hat points outward from P in the same direction as the line. Another arrow labeled vector E subscript out originates from the tip of the first arrow and points in the same direction.

Figure 2.3.10 A Gaussian surface surrounding a cylindrical shell.

Hence, the electric field at a point $P$ outside the shell at a distance $r$ away from the axis is

$\vec{\mathbf{E}}=\frac{2\pi R\sigma_0}{2\pi\epsilon_0}\frac{1}{r}\hat{\mathbf{r}}=\frac{R\sigma_0}{\epsilon_0}\frac{1}{r}\hat{\mathbf{r}}\ (r>R)$

where $\hat{\mathbf{r}}$ is a unit vector, perpendicular to the axis and pointing away from it, as shown in the figure. The electric field at $P$ points in the direction of $\hat{\mathbf{r}}$ given in Figure 2.3.10 if $\sigma_0>0$ and in the opposite direction to $\hat{\mathbf{r}}$ if $\sigma_0<0$ .

(b) Electric field at a point inside the shell. For a point inside the cylindrical shell, the Gaussian surface is a cylinder whose radius $r$ is less than $R$ (Figure 2.3.11). This means no charges are included inside the Gaussian surface:

$\lambda_{\mathrm{enc}}=0.$

(Figure 2.3.11) $\begin{gather*}.\end{gather*}$

Two cylinders sharing the same axis are shown. The inner one has length L, which is smaller than the outer cylinder’s length. An arrow labeled E subscript in originates from a point P on the inner cylinder and points outward, perpendicular to the axis.

Figure 2.3.11 A Gaussian surface within a cylindrical shell.

This gives the following equation for the magnitude of the electric field $E_{\mathrm{in}}$ at a point whose $r$ is less than $R$ of the shell of charges.

$E_{\mathrm{in}2\pi rL=0\ (r<R),$

This gives us

$E_{\mathrm{in}}=0\ (r<R).$

Significance

Notice that the result inside the shell is exactly what we should expect: No enclosed charge means zero electric field. Outside the shell, the result becomes identical to a wire with uniform charge $R\sigma_0$ .

CHECK YOUR UNDERSTANDING 2.5

A thin straight wire has a uniform linear charge density $\lambda_0$ . Find the electric field at a distance $d$ from the wire, where $d$ is much less than the length of the wire.

Charge Distribution with Planar Symmetry

A planar symmetry of charge density is obtained when charges are uniformly spread over a large flat surface. In planar symmetry, all points in a plane parallel to the plane of charge are identical with respect to the charges.

Consequences of symmetry

We take the plane of the charge distribution to be the $xy$ -plane and we find the electric field at a space point $P$ with coordinates $(x,y,z)$ . Since the charge density is the same at all $(x,y)$ -coordinates in the $z=0$ plane, by symmetry, the electric field at $P$ cannot depend on the $x$ – or $y$ -coordinates of point $P$ , as shown in Figure 2.3.12. Therefore, the electric field at $P$ can only depend on the distance from the plane and has a direction either toward the plane or away from the plane. That is, the electric field at $P$ has only a nonzero $z$ -component.

Uniform charges in $xy$ plane: $\vec{\mathbf{E}}=E(z)\hat{\mathbf{z}}$

where $z$ is the distance from the plane and $\hat{\mathbf{z}}$ is the unit vector normal to the plane. Note that in this system, $E(z)=E(-z)$ , although of course they point in opposite directions.

(Figure 2.3.12) $\begin{gather*}.\end{gather*}$

Figure shows a plane. Points q1 and q2 are on the plane, equidistant from its center. Lines connect these points to a point P above the plane. Arrows labeled vector E1 and vector E2 originate from point P and point in directions opposite to the lines connecting P to q1 and q2 respectively. A third arrow from P bisects the angle made by the first two arrows. This is labeled vector E subscript net.

Figure 2.3.12 The components of the electric field parallel to a plane of charges cancel out the two charges located symmetrically from the field point $P$ . Therefore, the field at any point is pointed vertically from the plane of charges. For any point $P$ and charge $q_1$ , we can always find a $q_2$ with this effect.

Gaussian surface and flux calculation

In the present case, a convenient Gaussian surface is a box, since the expected electric field points in one direction only. To keep the Gaussian box symmetrical about the plane of charges, we take it to straddle the plane of the charges, such that one face containing the field point $P$ is taken parallel to the plane of the charges. In Figure 2.3.13, sides I and II of the Gaussian surface (the box) that are parallel to the infinite plane have been shaded. They are the only surfaces that give rise to nonzero flux because the electric field and the area vectors of the other faces are perpendicular to each other.

(Figure 2.3.13) $\begin{gather*}.\end{gather*}$

Figure shows a cuboid and a plane through its center. The top and bottom surfaces of the cuboid are parallel to the plane and are labeled Slide 1 and Slide 2 respectively. An arrow labeled vector E subscript P originates from point P at the center of the top surface and points upwards, perpendicular to the surface. Another arrow labeled delta vector A also points up from the top surface. Two arrows labeled vector E and delta vector A point downward from the bottom surface. An arrow delta vector A originates from the right surface and points outward, perpendicular to the surface. Another arrow originates from its base. It is labeled vector E and points up.

Figure 2.3.13 A thin charged sheet and the Gaussian box for finding the electric field at the field point $P$ . The normal to each face of the box is from inside the box to outside. On two faces of the box, the electric fields are parallel to the area vectors, and on the other four faces, the electric fields are perpendicular to the area vectors.

Let $A$ be the area of the shaded surface on each side of the plane and $E_P$ be the magnitude of the electric field at point $P$ . Since sides I and II are at the same distance from the plane, the electric field has the same magnitude at points in these planes, although the directions of the electric field at these points in the two planes are opposite to each other.

Magnitude at I or II: $E(z)=E_P$ .

If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown in Figure 2.3.13. Therefore, we find for the flux of electric field through the box

(2.3.6) $\begin{equation*}\Phi=\oint_S\vec{\mathbf{E}}_P\cdot\hat{\mathbf{n}}dA=E_PA+E_PA+0+0+0+0=2E_PA,\end{equation*}$

where the zeros are for the flux through the other sides of the box. Note that if the charge on the plane is negative, the directions of electric field and area vectors for planes I and II are opposite to each other, and we get a negative sign for the flux. According to Gauss’s law, the flux must equal $q_{\mathrm{enc}}/\epsilon_0$ . From Figure 2.3.13, we see that the charges inside the volume enclosed by the Gaussian box reside on an area $A$ of the $xy$ -plane. Hence,

(2.3.7) $\begin{equation*}q_{\mathrm{enc}}=\sigma_0A.\end{equation*}$

Using the equations for the flux and enclosed charge in Gauss’s law, we can immediately determine the electric field at a point at height $z$ from a uniformly charged plane in the $xy$ -plane:

$\vec{\mathbf{E}}_P=\frac{\sigma_0}{2\epsilon_0}\hat{\mathbf{n}}.$

The direction of the field depends on the sign of the charge on the plane and the side of the plane where the field point $P$ is located. Note that above the plane, $\hat{\mathbf{n}}=+\hat{\mathbf{z}}$ , while below the plane, $\hat{\mathbf{n}}=-\hat{\mathbf{z}}$ .

You may be surprised to note that the electric field does not actually depend on the distance from the plane; this is an effect of the assumption that the plane is infinite. In practical terms, the result given above is still a useful approximation for finite planes near the centre.

14

2.4 Conductors in Electrostatic Equilibrium

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Describe the electric field within a conductor at equilibrium
Describe the electric field immediately outside the surface of a charged conductor at equilibrium
Explain why if the field is not as described in the first two objectives, the conductor is not at equilibrium

So far, we have generally been working with charges occupying a volume within an insulator. We now study what happens when free charges are placed on a conductor. Generally, in the presence of a (generally external) electric field, the free charge in a conductor redistributes and very quickly reaches electrostatic equilibrium. The resulting charge distribution and its electric field have many interesting properties, which we can investigate with the help of Gauss’s law and the concept of electric potential.

The Electric Field inside a Conductor Vanishes

If an electric field is present inside a conductor, it exerts forces on the free electrons (also called conduction electrons), which are electrons in the material that are not bound to an atom. These free electrons then accelerate. However, moving charges by definition means nonstatic conditions, contrary to our assumption. Therefore, when electrostatic equilibrium is reached, the charge is distributed in such a way that the electric field inside the conductor vanishes.

If you place a piece of a metal near a positive charge, the free electrons in the metal are attracted to the external positive charge and migrate freely toward that region. The region the electrons move to then has an excess of electrons over the protons in the atoms and the region from where the electrons have migrated has more protons than electrons. Consequently, the metal develops a negative region near the charge and a positive region at the far end (Figure 2.4.1). As we saw in the preceding chapter, this separation of equal magnitude and opposite type of electric charge is called polarization. If you remove the external charge, the electrons migrate back and neutralize the positive region.

(Figure 2.4.1) $\begin{gather*}.\end{gather*}$

Figure shows a sphere and a positive charge q some distance away from it. The side of the sphere facing q is labeled A and the opposite side is labeled B. Minus signs and plus signs are shown at the inner surfaces of the sphere on sides A and B respectively. These are labeled minus sigma A and plus sigma B respectively.

Figure 2.4.1 Polarization of a metallic sphere by an external point charge $+q$ . The near side of the metal has an opposite surface charge compared to the far side of the metal. The sphere is said to be polarized. When you remove the external charge, the polarization of the metal also disappears.

The polarization of the metal happens only in the presence of external charges. You can think of this in terms of electric fields. The external charge creates an external electric field. When the metal is placed in the region of this electric field, the electrons and protons of the metal experience electric forces due to this external electric field, but only the conduction electrons are free to move in the metal over macroscopic distances. The movement of the conduction electrons leads to the polarization, which creates an induced electric field in addition to the external electric field (Figure 2.4.2). The net electric field is a vector sum of the fields of $+q$ and the surface charge densities $-\sigma_A$ and $+\sigma_B$ This means that the net field inside the conductor is different from the field outside the conductor.

(Figure 2.4.2) $\begin{gather*}.\end{gather*}$

Figure shows a sphere and a charge q some distance away from it. The side of the sphere facing q is labeled A and the opposite side is labeled B. The inner surfaces of the sphere on sides A and B are labeled minus sigma A and plus sigma B respectively. A point P is on the sphere. Two arrows originate from P. They are labeled vector E subscript A and vector E subscript B. A dotted line bisects the angle formed by the two and connects P to q. A third arrow originates from P and points in the direction opposite to q. This is labeled vector E subscript q.

Figure 2.4.2 In the presence of an external charge $q$ , the charges in a metal redistribute. The electric field at any point has three contributions, from $+q$ and the induced charges $-\sigma_A$ and $+\sigma_B$ . Note that the surface charge distribution will not be uniform in this case.

The redistribution of charges is such that the sum of the three contributions at any point $P$ inside the conductor is

$\vec{\mathbf{E}}_P=\vec{\mathbf{E}}_q+\vec{\mathbf{E}}_B+\vec{\mathbf{E}}_A=\vec{\mathbf{0}}.$

Now, thanks to Gauss’s law, we know that there is no net charge enclosed by a Gaussian surface that is solely within the volume of the conductor at equilibrium. That is, $q_{\mathrm{enc}}=0$ and hence

(2.4.1) $\begin{equation*}\vec{\mathbf{E}}_{\mathrm{net}}=\vec{\mathbf{0}}\ (\mathrm{at~points~inside~a~conductor}).\end{equation*}$

Charge on a Conductor

An interesting property of a conductor in static equilibrium is that extra charges on the conductor end up on the outer surface of the conductor, regardless of where they originate. Figure 2.4.3 illustrates a system in which we bring an external positive charge inside the cavity of a metal and then touch it to the inside surface. Initially, the inside surface of the cavity is negatively charged and the outside surface of the conductor is positively charged. When we touch the inside surface of the cavity, the induced charge is neutralized, leaving the outside surface and the whole metal charged with a net positive charge.

(Figure 2.4.3) $\begin{gather*}.\end{gather*}$

A figure on the left shows a shaded circle with a cavity in it. A rod with a ball at the end is inserted in the cavity in such a way that it does not touch the shaded circle. The ball has a plus sign on it. The cavity has minus signs around it. The shaded circle has plus signs outside it. An arrow points from this figure to a figure on the right. The arrow is labeled touch inside cavity. The figure on the right is similar to the figure on the left, except that the ball is touching the edge of the cavity. There are no signs on the ball or around the cavity. The outside of the shaded circle has plus signs.

Figure 2.4.3 Electric charges on a conductor migrate to the outside surface no matter where you put them initially.

To see why this happens, note that the Gaussian surface in Figure 2.4.4 (the dashed line) follows the contour of the actual surface of the conductor and is located an infinitesimal distance within it. Since $E=0$ everywhere inside a conductor,

$\oing_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=0.$

Thus, from Gauss’ law, there is no net charge inside the Gaussian surface. But the Gaussian surface lies just below the actual surface of the conductor; consequently, there is no net charge inside the conductor. Any excess charge must lie on its surface.

(Figure 2.4.4) $\begin{gather*}.\end{gather*}$

Figure shows an irregular shape. A dotted line is shown just inside the outline of the shape.

Figure 2.4.4 The dashed line represents a Gaussian surface that is just beneath the actual surface of the conductor.

This particular property of conductors is the basis for an extremely accurate method developed by Plimpton and Lawton in 1936 to verify Gauss’s law and, correspondingly, Coulomb’s law. A sketch of their apparatus is shown in Figure 2.4.5. Two spherical shells are connected to one another through an electrometer $\mathrm{E}$ , a device that can detect a very slight amount of charge flowing from one shell to the other. When switch $\mathrm{S}$ is thrown to the left, charge is placed on the outer shell by the battery $\mathrm{B}$ . Will charge flow through the electrometer to the inner shell?

No. Doing so would mean a violation of Gauss’s law. Plimpton and Lawton (1936) did not detect any flow and, knowing the sensitivity of their electrometer, concluded that if the radial dependence in Coulomb’s law were $1/r^{2+\delta}$ , $\delta$ would be less than $2\times10^{-9}$ . More recent measurements by Williams, Faller, and Hill (1971) place $\delta$ at less than $3\times10^{-16}$ , a number so small that the validity of Coulomb’s law seems indisputable.

(Figure 2.4.5) $\begin{gather*}.\end{gather*}$

Figure shows a circle labeled E. It is surrounded by two concentric circles with slits in them. These are labeled two concentric conducting spheres. Two terminals of E are connected, one to each circle. The outer circle is connected to a switch S, which switches between two terminals of a battery. There is a slanted plate at the top of the circles labeled mirror for viewing the electrometer. A person views the mirror through a scope. The line of view is reflected from the mirror to E.

Figure 2.4.5 A representation of the apparatus used by Plimpton and Lawton. Any transfer of charge between the spheres is detected by the electrometer $\mathrm{E}$ .

The Electric Field at the Surface of a Conductor

If the electric field had a component parallel to the surface of a conductor, free charges on the surface would move, a situation contrary to the assumption of electrostatic equilibrium. Therefore, the electric field is always perpendicular to the surface of a conductor.

At any point just above the surface of a conductor, the surface charge density $\delta$ and the magnitude of the electric field $E$ are related by

(2.4.2) $\begin{equation*}E=\frac{\sigma}{\epsilon_0}.\end{equation*}$

To see this, consider an infinitesimally small Gaussian cylinder that surrounds a point on the surface of the conductor, as in Figure 2.4.6. The cylinder has one end face inside and one end face outside the surface. The height and cross-sectional area of the cylinder are $\delta$ and $\Delta A$ , respectively. The cylinder’s sides are perpendicular to the surface of the conductor, and its end faces are parallel to the surface. Because the cylinder is infinitesimally small, the charge density $\sigma$ is essentially constant over the surface enclosed, so the total charge inside the Gaussian cylinder is $\sigma\Delta A$ . Now $E$ is perpendicular to the surface of the conductor outside the conductor and vanishes within it, because otherwise, the charges would accelerate, and we would not be in equilibrium. Electric flux therefore crosses only the outer end face of the Gaussian surface and may be written as $E\Delta A$ , since the cylinder is assumed to be small enough that $E$ is approximately constant over that area. From Gauss’ law,

$E\Delta A=\frac{\sigma\Delta A}{\epsilon_0}.$

Thus,

$E=\frac{\sigma}{\epsilon_0}.$

(Figure 2.4.6) $\begin{gather*}.\end{gather*}$

A surface labeled sigma has plus signs on it. A point P on the surface forms the center of a cylinder. An arrow labeled vector E is along the axis of the cylinder and emerges from its top surface. The top surface of the cylinder is labeled delta A and the bottom surface is labeled vector E equal to zero. These are parallel to the surface sigma. The length of the cylinder is labeled sigma.

Figure 2.4.6 An infinitesimally small cylindrical Gaussian surface surrounds point $P$ , which is on the surface of the conductor. The field $\vec{\mathbf{E}}$ is perpendicular to the surface of the conductor outside the conductor and vanishes within it.

EXAMPLE 2.4.1

Electric Field of a Conducting Plate

The infinite conducting plate in Figure 2.4.7 has a uniform surface charge density $\sigma$ . Use Gauss’ law to find the electric field outside the plate. Compare this result with that previously calculated directly.

(Figure 2.4.7) $\begin{gather*}.\end{gather*}$

A shaded strip labeled E equal to zero has plus signs on both its inner edges. A rectangle labeled A is shown on the right of the strip such that it encloses two plus signs. Two arrows within this are perpendicular to the length of the strip and point right. These are labeled vector E.

Figure 2.4.7 A side view of an infinite conducting plate and Gaussian cylinder with cross-sectional area $A$ .

Strategy

For this case, we use a cylindrical Gaussian surface, a side view of which is shown.

Solution

The flux calculation is similar to that for an infinite sheet of charge from the previous chapter with one major exception: The left face of the Gaussian surface is inside the conductor where $\vec{\mathbf{E}}=\vec{\mathbf{0}}$ , so the total flux through the Gaussian surface is $EA$ rather than $2EA$ . Then from Gauss’ law,

$EA=\frac{\sigma A}{\epsilon_0}$

and the electric field outside the plate is

$E=\frac{\sigma}{\epsilon_0}.$

Significance

This result is in agreement with the result from the previous section, and consistent with the rule stated above.

EXAMPLE 2.4.2

Electric Field between Oppositely Charged Parallel Plates

Two large conducting plates carry equal and opposite charges, with a surface charge density $\sigma$ of magnitude $6.81\times10^{-7}~\mathrm{C/m}^2$ , as shown in Figure 2.4.8. The separation between the plates is $l=6.50~\mathrm{mm}$ . What is the electric field between the plates?

(Figure 2.4.8) $\begin{gather*}.\end{gather*}$

Two parallel plates are shown, a distance l apart. The left one has plus signs on the inside of its right surface. the right plate has minus signs on the inside of its left surface. Arrows from the left plate to the right plate are labeled vector E. A positive charge between the plates has an arrow from it, pointing right.

Figure 2.4.8 The electric field between oppositely charged parallel plates. A test charge is released at the positive plate.

Strategy

Note that the electric field at the surface of one plate only depends on the charge on that plate. Thus, apply $E=\sigma/\epsilon_0$ with the given values.

Solution

The electric field is directed from the positive to the negative plate, as shown in the figure, and its magnitude is given by

$E=\frac{\sigma}{\epsilon_0}=\frac{6.81\times10^{-7}~\mathrm{C/m}^2}{8.85\times10^{-12}~\mathrm{C}^2/\mathrm{N~m}^2}=7.69\times10^4~\mathrm{N/C}.$

Significance

This formula is applicable to more than just a plate. Furthermore, two-plate systems will be important later.

EXAMPLE 2.4.3

A Conducting Sphere

The isolated conducting sphere (Figure 2.4.9) has a radius $R$ and an excess charge $q$ . What is the electric field both inside and outside the sphere?

(Figure 2.4.9) $\begin{gather*}.\end{gather*}$

Two concentric circles are shown. The smaller one, with radius R, has plus signs around the inside of it. The bigger one, with radius r is shown with a dotted line and labeled S, Gaussian surface.

Figure 2.4.9 An isolated conducting sphere.

Strategy

The sphere is isolated, so its surface change distribution and the electric field of that distribution are spherically symmetrical. We can therefore represent the field as $\vec{\mathbf{E}}=E(r)\hat{\mathbf{r}}$ . To calculate $E(r)$ , we apply Gauss’s law over a closed spherical surface $S$ of radius $r$ that is concentric with the conducting sphere.

Solution

Since $r$ is constant and $\hat{\mathbf{n}}=\hat{\mathbf{r}}$ on the sphere,

$\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=E(r)\oint_SdA=E(r)4\pi r^2.$

r<R

$E(r)=0,$

as expected inside a conductor. If $r>R$ , $S$ encloses the conductor so $q_{\mathrm{enc}}$ . From Gauss’s law,

$E(r)4\pi r^2=\frac{q}{\epsilon_0}.$

The electric field of the sphere may therefore be written as

$\begin{array}{ll}\vec{\mathbf{E}}=\vec{\mathbf{0}}&(r<R),\\\vec{\mathbf{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{\mathbf{r}}&(r\geq R).\end{array}$

Significance

Notice that in the region $r\geq0$ , the electric field due to a charge $q$ placed on an isolated conducting sphere of radius $R$ is identical to the electric field of a point charge $q$ located at the centre of the sphere. The difference between the charged metal and a point charge occurs only at the space points inside the conductor. For a point charge placed at the centre of the sphere, the electric field is not zero at points of space occupied by the sphere, but a conductor with the same amount of charge has a zero electric field at those points (Figure 2.4.10). However, there is no distinction at the outside points in space where $r>R$ , and we can replace the isolated charged spherical conductor by a point charge at its centre with impunity.

(Figure 2.4.10) $\begin{gather*}.\end{gather*}$

A circle labeled vector E subscript in equal to zero is shown. Arrows around it radiate outwards. These are labeled vector E subscript out.

Figure 2.4.10 Electric field of a positively charged metal sphere. The electric field inside is zero, and the electric field outside is same as the electric field of a point charge at the center, although the charge on the metal sphere is at the surface.

CHECK YOUR UNDERSTANDING 2.6

How will the system above change if there are charged objects external to the sphere?

For a conductor with a cavity, if we put a charge $+q$ inside the cavity, then the charge separation takes place in the conductor, with $-q$ amount of charge on the inside surface and a $+q$ amount of charge at the outside surface (Figure 2.4.11(a)). For the same conductor with a charge $+q$ outside it, there is no excess charge on the inside surface; both the positive and negative induced charges reside on the outside surface (Figure 2.4.11(b)).

(Figure 2.4.11) $\begin{gather*}.\end{gather*}$

Figure a shows a metal sphere with a cavity within it. The sphere is labeled vector E equal to zero. It has plus signs around it. The cavity has minus signs around it. A positive charge plus q is within the cavity. Figure b shows the same metal sphere with a cavity in it. The sphere is labeled vector E equal to zero. There is nothing within the cavity. A positive charge labeled plus q is outside the sphere. The side of the sphere facing q has minus signs on it. The opposite side has plus signs on it.

Figure 2.4.11 (a) A charge inside a cavity in a metal. The distribution of charges at the outer surface does not depend on how the charges are distributed at the inner surface, since the $E$ -field inside the body of the metal is zero. That magnitude of the charge on the outer surface does depend on the magnitude of the charge inside, however. (b) A charge outside a conductor containing an inner cavity. The cavity remains free of charge. The polarization of charges on the conductor happens at the surface.

If a conductor has two cavities, one of them having a charge $+q_a$ inside it and the other a charge $-q_b$ , the polarization of the conductor results in $-q_a$ on the inside surface of the cavity $a$ , $+q_b$ on the inside surface of the cavity $b$ , and $q_a-q_b$ on the outside surface (Figure 2.4.12). The charges on the surfaces may not be uniformly spread out; their spread depends upon the geometry. The only rule obeyed is that when the equilibrium has been reached, the charge distribution in a conductor is such that the electric field by the charge distribution in the conductor cancels the electric field of the external charges at all space points inside the body of the conductor.

(Figure 2.4.12) $\begin{gather*}.\end{gather*}$

Figure shows a flattened sphere, labeled vector E equal to zero. It has two spherical cavities within it. Its outer surface of the flattened sphere is labeled no induced charge outside. The left cavity has a negative charge q inside it, on the left. The left surface of this cavity has many plus signs on it and the right surface has a single plus sign on it. The right cavity has a positive charge q inside it, on the right. The right surface of this cavity has many minus signs on it and the left surface has a single minus sign on it.

Figure 2.4.12 The charges induced by two equal and opposite charges in two separate cavities of a conductor. If the net charge on the cavity is nonzero, the external surface becomes charged to the amount of the net charge.

15

Chapter 2 Review

Key Terms

area vector
vector with magnitude equal to the area of a surface and direction perpendicular to the surface

cylindrical symmetry
system only varies with distance from the axis, not direction

electric flux
dot product of the electric field and the area through which it is passing

flux
quantity of something passing through a given area

free electrons
also called conduction electrons, these are the electrons in a conductor that are not bound to any particular atom, and hence are free to move around

Gaussian surface
any enclosed (usually imaginary) surface

planar symmetry
system only varies with distance from a plane

spherical symmetry
system only varies with the distance from the origin, not in direction

Key Equations

Definition of electric flux, for uniform electric field	$\Phi=\vec{\mathbf{E}}\cdot\vec{\mathbf{A}}\rightarrow EA\cos\theta$
Electric flux through an open surface	$\Phi=\int_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\int_S\vec{\mathbf{E}}\cdot d\vec{\mathbf{A}}$
Electric flux through a closed surface	$\Phi=\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\oint_S\vec{\mathbf{E}}\cdot\vec{\mathbf{E}}\cdot d\vec{\mathbf{A}}$
Gauss’s law	$\Phi=\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\frac{q_{\mathrm{enc}}}{\epsilon_0}$
Gauss’s Law for systems with symmetry	$\Phi=\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=E\oint_SdA=EA=\frac{q_{\mathrm{enc}}}{\epsilon_0}$
The magnitude of the electric field just outside the surface of a conductor	$E=\frac{\sigma}{\epsilon_0}$

Summary

2.1 Electric Flux

The electric flux through a surface is proportional to the number of field lines crossing that surface. Note that this means the magnitude is proportional to the portion of the field perpendicular to the area.
The electric flux is obtained by evaluating the surface integral
$\Phi=\int_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\int_S\vec{\mathbf{E}}\cdot d\vec{\mathbf{A}}$

where the notation used here is for a closed surface $S$ .

2.2 Explaining Gauss’s Law

Gauss’s law relates the electric flux through a closed surface to the net charge within that surface,
$\Phi=\oint_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA=\frac{q_{\mathrm{enc}}}{\epsilon_0}$

where qenc is the total charge inside the Gaussian surface $S$ .
All surfaces that include the same amount of charge have the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surfaces enclose the same amount of charge.

2.3 Applying Gauss’s Law

For a charge distribution with certain spatial symmetries (spherical, cylindrical, and planar), we can find a Gaussian surface over which $\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}=E$ , where $E$ is constant over the surface. The electric field is then determined with Gauss’s law.
For spherical symmetry, the Gaussian surface is also a sphere, and Gauss’s law simplifies to $4\pi r^2E=\frac{q_{\mathrm{enc}}}{\epsilon_0}$ .
For cylindrical symmetry, we use a cylindrical Gaussian surface, and find that Gauss’s law simplifies to $2\pi rLE=\frac{q_{\mathrm{enc}}}{\epsilon_0}$ .
For planar symmetry, a convenient Gaussian surface is a box penetrating the plane, with two faces parallel to the plane and the remainder perpendicular, resulting in Gauss’s law being $2AE=\frac{q_{\mathrm{enc}}}{\epsilon_0}$ .

2.4 Conductors in Electrostatic Equilibrium

The electric field inside a conductor vanishes.
Any excess charge placed on a conductor resides entirely on the surface of the conductor.
The electric field is perpendicular to the surface of a conductor everywhere on that surface.
The magnitude of the electric field just above the surface of a conductor is given by $E=\frac{\sigma}{\epsilon_0}$ .

Answers to Check Your Understanding

2.1. Place it so that its unit normal is perpendicular to $\vec{\mathbf{E}}$ .

2.2. $mab^2/2$ .

2.3 a. $3.4\times10^5~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ ; b. $-3.4\times10^5~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ ; c. $3.4\times10^5~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ ; d. $0$ .

2.4. In this case, there is only $\vec{\mathbf{E}}_{\mathrm{out}}$ . So, yes.

2.5. $\vec{\mathbf{E}}=\frac{\lambda_0}{2\pi\epsilon_0}\frac{1}{d}\hat{\mathbf{r}}$ ; This agrees with the calculation of Example 1.5.1 where we found the electric field by integrating over the charged wire. Notice how much simpler the calculation of this electric field is with Gauss’s law.

2.6. If there are other charged objects around, then the charges on the surface of the sphere will not necessarily be spherically symmetrical; there will be more in certain direction than in other directions.

Conceptual Questions

2.1 Electric Flux

1. Discuss how would orient a planar surface of area $A$ in a uniform electric field of magnitude $E_0$ to obtain (a) the maximum flux and (b) the minimum flux through the area.

2. What are the maximum and minimum values of the flux in the preceding question?

3. The net electric flux crossing a closed surface is always zero. True or false?

4. The net electric flux crossing an open surface is never zero. True or false?

2.2 Explaining Gauss’s Law

5. Two concentric spherical surfaces enclose a point charge $q$ . The radius of the outer sphere is twice that of the inner one. Compare the electric fluxes crossing the two surfaces.

6. Compare the electric flux through the surface of a cube of side length $a$ that has a charge $q$ at its centre to the flux through a spherical surface of radius a with a charge $q$ at its centre.

7. (a) If the electric flux through a closed surface is zero, is the electric field necessarily zero at all points on the surface? (b) What is the net charge inside the surface?

8. Discuss how Gauss’s law would be affected if the electric field of a point charge did not vary as $1/r^2$ .

9. Discuss the similarities and differences between the gravitational field of a point mass m and the electric field of a point charge $q$ .

10. Discuss whether Gauss’s law can be applied to other forces, and if so, which ones.

11. Is the term $\vec{\mathbf{E}}$ in Gauss’s law the electric field produced by just the charge inside the Gaussian surface?

12. Reformulate Gauss’s law by choosing the unit normal of the Gaussian surface to be the one directed inward.

2.3 Applying Gauss’s Law

13. Would Gauss’s law be helpful for determining the electric field of two equal but opposite charges a fixed distance apart?

14. Discuss the role that symmetry plays in the application of Gauss’s law. Give examples of continuous charge distributions in which Gauss’s law is useful and not useful in determining the electric field.

15. Discuss the restrictions on the Gaussian surface used to discuss planar symmetry. For example, is its length important? Does the cross-section have to be square? Must the end faces be on opposite sides of the sheet?

2.4 Conductors in Electrostatic Equilibrium

16. Is the electric field inside a metal always zero?

17. Under electrostatic conditions, the excess charge on a conductor resides on its surface. Does this mean that all the conduction electrons in a conductor are on the surface?

18. A charge $q$ is placed in the cavity of a conductor as shown below. Will a charge outside the conductor experience an electric field due to the presence of $q$ ?

Figure shows an egg shape with an oval cavity within it. The cavity is surrounded by a dotted line just outside it. This is labeled S. There is a positive charge labeled q within the cavity.

19. The conductor in the preceding figure has an excess charge of $-5.0~\mu\mathrm{C}$ . If a $2.0{\text -}\mu\mathrm{C}$ point charge is placed in the cavity, what is the net charge on the surface of the cavity and on the outer surface of the conductor?

Problems

2.1 Electric Flux

20. A uniform electric field of magnitude $1.1\times10^4~\mathrm{N/C}$ is perpendicular to a square sheet with sides $2.0~\mathrm{m}$ long. What is the electric flux through the sheet?

21. Calculate the flux through the sheet of the previous problem if the plane of the sheet is at an angle of $60^{\circ}$ to the field. Find the flux for both directions of the unit normal to the sheet.

22. Find the electric flux through a rectangular area $3~\mathrm{cm}\times2~\mathrm{cm}$ between two parallel plates where there is a constant electric field of $30~\mathrm{N/C}$ for the following orientations of the area: (a) parallel to the plates, (b) perpendicular to the plates, and (c) the normal to the area making a $30^{\circ}$ angle with the direction of the electric field. Note that this angle can also be given as $180^{\circ}+30^{\circ}$ .

23. The electric flux through a square-shaped area of side $5~\mathrm{cm}$ near a large charged sheet is found to be $3\times10^{-5}~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ when the area is parallel to the plate. Find the charge density on the sheet.

24. Two large rectangular aluminum plates of area $150~\mathrm{cm}^2$ face each other with a separation of $3~\mathrm{mm}$ between them. The plates are charged with equal amount of opposite charges, $\pm20~\mu\mahtrm{C}$ . The charges on the plates face each other. Find the flux through a circle of radius $3~\mathrm{cm}$ between the plates when the normal to the circle makes an angle of $5^{\circ}$ with a line perpendicular to the plates. Note that this angle can also be given as $180^{\circ}+5^{\circ}$ .

25. A square surface of area $2~\mathrm{cm}^2$ is in a space of uniform electric field of magnitude $10^3~\mathrm{N/C}$ . The amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when the normal to it makes the following angles with electric field: (a) $30^{\circ}$ , (b) $90^{\circ}$ and (c) $0^{\circ}$ . Note that these angles can also be given as $180^{\circ}+\theta$ .

26. A vector field is pointed along the $z$ -axis, $\vec{\mathbf{v}}=\frac{\alpha}{x^2+y^2}\hat{\mathbf{z}}$ . (a) Find the flux of the vector field through a rectangle in the $xy$ -plane between $a<x<b$ and $c<y<d$ . (b) Do the same through a rectangle in the $yz$ -plane between $a<z<b$ and $c<y<d$ . (Leave your answer as an integral.)

27. Consider the uniform electric field $\vec{\mathbf{E}}=(4.0\hat{\mathbf{j}}+3.0\hat{\mathbf{k}})\times10^3~\mathrm{N/C}$ . What is its electric flux through a circular area of radius $2.0~\mathrm{m}$ that lies in the $xy$ -plane?

28. Repeat the previous problem, given that the circular area is (a) in the $yz$ -plane and (b) $45^{\circ}$ above the $xy$ –plane.

29. An infinite charged wire with charge per unit length $\lambda$ lies along the central axis of a cylindrical surface of radius $r$ and length $l$ . What is the flux through the surface due to the electric field of the charged wire?

2.2 Explaining Gauss’s Law

30. Determine the electric flux through each surface whose cross-section is shown below.

31. Find the electric flux through the closed surface whose cross-sections are shown below.

32. A point charge $q$ is located at the centre of a cube whose sides are of length $a$ . If there are no other charges in this system, what is the electric flux through one face of the cube?

33. A point charge of $10~\mu\mathrm{C}$ is at an unspecified location inside a cube of side $2~\mathrm{cm}$ . Find the net electric flux though the surfaces of the cube.

34. A net flux of $1.0\times10^4~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ passes inward through the surface of a sphere of radius $5~\mathrm{cm}$ . (a) How much charge is inside the sphere? (b) How precisely can we determine the location of the charge from this information?

35. A charge $q$ is placed at one of the corners of a cube of side $a$ , as shown below. Find the magnitude of the electric flux through the shaded face due to $q$ . Assume $a>0$ .

Figure shows a cube with length of each side equal to a. The back surface of it is shaded. One front corner has a small circle on it labeled q.

36. The electric flux through a cubical box $8.0~\mathrm{cm}$ on a side is $1.2\times10^3~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ . What is the total charge enclosed by the box?

37. The electric flux through a spherical surface is $4.0\times10^4~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}$ . What is the net charge enclosed by the surface?

38. A cube whose sides are of length $d$ is placed in a uniform electric field of magnitude $E=4.0\times10^3~\mathrm{N/C}$ so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube?

39. Repeat the previous problem, assuming that the electric field is directed along a body diagonal of the cube.

40. A total charge $5.0\times10^{6}~\mathrm{C}$ is distributed uniformly throughout a cubical volume whose edges are $8.0~\mathrm{cm}$ long. (a) What is the charge density in the cube? (b) What is the electric flux through a cube with $12.0{\text -}\mathrm{cm}$ edges that is concentric with the charge distribution? (c) Do the same calculation for cubes whose edges are $10.0~\mathrm{cm}$ long and $5.0~\mathrm{cm}$ long. (d) What is the electric flux through a spherical surface of radius $3.0~\mathrm{cm}$ that is also concentric with the charge distribution?

2.3 Applying Gauss’s Law

41. Recall that in the example of a uniform charged sphere, $\rho_0=Q/\left(\frac{4}{3}\pi R^3\right)$ . Rewrite the answers in terms of the total charge $Q$ on the sphere.

42. Suppose that the charge density of the spherical charge distribution shown in Figure 2.3.3 is $\rho(r)=\rho_0r/R$ for $r\leq R$ and zero for $r>R$ . Obtain expressions for the electric field both inside and outside the distribution.

43. A very long, thin wire has a uniform linear charge density of $50~\mu\mathrm{C/m}$ . What is the electric field at a distance $2.0~\mathrm{cm}$ from the wire?

44. A charge of $-30~\mu\mathrm{C}$ is distributed uniformly throughout a spherical volume of radius $10.0~\mathrm{cm}$ . Determine the electric field due to this charge at a distance of (a) $2.0~\mathrm{cm}$ , (b) $5.0~\mathrm{cm}$ , and (c) $20.0~\mathrm{cm}$ from the centre of the sphere.

45. Repeat your calculations for the preceding problem, given that the charge is distributed uniformly over the surface of a spherical conductor of radius $10.0~\mathrm{cm}$ .

46. A total charge $Q$ is distributed uniformly throughout a spherical shell of inner and outer radii $r_1$ and $r_2$ , respectively. Show that the electric field due to the charge is

$\begin{array}{ll}\vec{\mathbf{E}}=\vec{\mathbf{0}}&(r\leqr_1);\\\vec{\mathbf{E}}=\frac{Q}{4\pi\epsilon_0r^2}\left(\frac{r^3-r_1^3}{r_2^3-r_1^3}\right)\hat{\mathbf{r}}&(r_1\leq r\leq r_2);\\\vec{\mathbf{E}}=\frac{Q}{4\pi\epsilon_0r^2}\hat{\mathbf{r}}&(r\geq r_2).\end{array}$

47. When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of $+3.0~\mu\mathrm{C}$ charge put on a $5.0{\text -}\mathrm{cm}$ aluminum spherical ball at the following two points in space: (a) a point $1.0~\mathrm{cm}$ from the centre of the ball (an inside point) and (b) a point $10~\mathrm{cm}$ from the centre of the ball (an outside point).

48. A large sheet of charge has a uniform charge density of $10~\mu\mathrm{C/m}^2$ . What is the electric field due to this charge at a point just above the surface of the sheet?

49. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A $300{\text -}\mathrm{cm}$ long copper rod of radius $1~\mathrm{cm}$ is charged with $+500~\mathrm{nC}$ of charge and we seek electric field at a point $5~\mathrm{cm}$ from the centre of the rod. (b) A $10{\text -}\mathrm{cm}$ long copper rod of radius $1~\mathrm{cm}$ is charged with $+500~\mathrm{nC}$ of charge and we seek electric field at a point $5~\mathrm{cm}$ from the centre of the rod. (c) A $150{\text -}\mathrm{cm}$ wooden rod is glued to a $150{\text -}\mathrm{cm}$ plastic rod to make a $300{\text -}\mathrm{cm}$ long rod, which is then painted with a charged paint so that one obtains a uniform charge density. The radius of each rod is $1~\mathrm{cm}$ , and we seek an electric field at a point that is $4~\mathrm{cm}$ from the centre of the rod. (d) Same rod as (c), but we seek electric field at a point that is $500~\mathrm{cm}$ from the centre of the rod.

50. A long silver rod of radius $3~\mathrm{cm}$ has a charge of $-5~\mu\mathrm{C/cm}$ on its surface. (a) Find the electric field at a point $5~\mathrm{cm}$ from the centre of the rod (an outside point). (b) Find the electric field at a point $2~\mathrm{cm}$ from the centre of the rod (an inside point).

51. The electric field at $2~\mathrm{cm}$ from the centre of long copper rod of radius $1~\mathrm{cm}$ has a magnitude $3~\mathrm{N/C}$ and directed outward from the axis of the rod. (a) How much charge per unit length exists on the copper rod? (b) What would be the electric flux through a cube of side $5~\mathrm{cm}$ situated such that the rod passes through opposite sides of the cube perpendicularly?

52. A long copper cylindrical shell of inner radius $2~\mathrm{cm}$ and outer radius $3~\mathrm{cm}$ surrounds concentrically a charged long aluminum rod of radius $1~\mathrm{cm}$ with a charge density of $4~\mathrm{pC/m}$ . All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the centre of the aluminum rod: (a) $0.5~\mathrm{cm}$ , (b) $1.5~\mathrm{cm}$ , (c) $2.5~\mathrm{cm}$ , (d) $3.5~\mathrm{cm}$ , and (e) $7\mathrm{cm}$ .

53. Charge is distributed uniformly with a density $\rho$ throughout an infinitely long cylindrical volume of radius $R$ . Show that the field of this charge distribution is directed radially with respect to the cylinder and that

$\begin{array}{ll}E=\frac{\rho r}{2\epsilon_0}&(r\leq R);\\~&~\\E=\frac{\rho R^2}{2\epsilon_0r}&(r\geq R).\end{array}$

54. Charge is distributed throughout a very long cylindrical volume of radius $R$ such that the charge density increases with the distance $r$ from the central axis of the cylinder according to $\rho=\alpha r$ , where $\alpha$ is a constant. Show that the field of this charge distribution is directed radially with respect to the cylinder and that

$\begin{array}{ll}E=\frac{\alpha r^2}{2\epsilon_0}&(r\leq R);\\~&~\\E=\frac{\alpha R^3}{3\epsilon_0r}&(r\geq R).\end{array}$

55. The electric field $10.0~\mathrm{cm}$ from the surface of a copper ball of radius $5.0~\mathrm{cm}$ is directed toward the ball’s centre and has magnitude $4.0\times10^2~\mathrm{N/C}$ . How much charge is on the surface of the ball?

56. Charge is distributed throughout a spherical shell of inner radius $r_1$ and outer radius $r_2$ with a volume density given by $\rho=\rho_0r_1/r$ , where $\rho_0$ is a constant. Determine the electric field due to this charge as a function of $r$ , the distance from the centre of the shell.

57. Charge is distributed throughout a spherical volume of radius $R$ with a density $\rho=\alpha r^2$ , where $\alpha$ is a constant. Determine the electric field due to the charge at points both inside and outside the sphere.

58. Consider a uranium nucleus to be sphere of radius $R=7.4\times10^{-15}~\mathrm{m}$ with a charge of $92e$ distributed uniformly throughout its volume. (a) What is the electric force exerted on an electron when it is $3.0\times10^{-15}~\mathrm{m}$ from the centre of the nucleus? (b) What is the acceleration of the electron at this point?

59. The volume charge density of a spherical charge distribution is given by $\rho(r)=\rho_0e^{-\alpha r}$ , where $\rho_0$ and $\alpha$ are constants. What is the electric field produced by this charge distribution?

2.4 Conductors in Electrostatic Equilibrium

60. An uncharged conductor with an internal cavity is shown in the following figure. Use the closed surface $S$ along with Gauss’ law to show that when a charge $q$ is placed in the cavity a total charge $-q$ is induced on the inner surface of the conductor. What is the charge on the outer surface of the conductor?

A metal sphere with a cavity is shown. It is labeled vector E equal to zero. There are plus signs surrounding it. There is a positive charge labeled plus q within the cavity. The cavity is surrounded by minus signs.

Figure 2.5.1 A charge inside a cavity of a metal. Charges at the outer surface do not depend on how the charges are distributed at the inner surface since $E$ field inside the body of the metal is zero.

61. An uncharged spherical conductor $S$ of radius $R$ has two spherical cavities $\mathrm{A}$ and $\mathrm{B}$ of radii $a$ and $b$ , respectively as shown below. Two point charges $+q_a$ and $+q_b$ are placed at the centre of the two cavities by using non-conducting supports. In addition, a point charge $+q_0$ is placed outside at a distance $r$ from the centre of the sphere. (a) Draw approximate charge distributions in the metal although metal sphere has no net charge. (b) Draw electric field lines. Draw enough lines to represent all distinctly different places.
Figure shows a sphere with two cavities. A positive charge qa is in one cavity and a positive charge qb is in the other cavity. A positive charge q0 is outside the sphere at a distance r from its center.

62. A positive point charge is placed at the angle bisector of two uncharged plane conductors that make an angle of $45^{\circ}$ . See below. Draw the electric field lines.
An acute angle is shown. Its bisector is a dotted line. A positive charge q is shown on the dotted line.

63. A long cylinder of copper of radius $3~\mathrm{cm}$ is charged so that it has a uniform charge per unit length on its surface of $3~\mathrm{C/m}$ . (a) Find the electric field inside and outside the cylinder. (b) Draw electric field lines in a plane perpendicular to the rod.

64. An aluminum spherical ball of radius $4~\mathrm{cm}$ is charged with $5~\mu\mathrm{C}$ of charge. A copper spherical shell of inner radius $6~\mathrm{cm}$ and outer radius $8~\mathrm{cm}$ surrounds it. A total charge of $-8~\mu\mathrm{C}$ is put on the copper shell. (a) Find the electric field at all points in space, including points inside the aluminum and copper shell when copper shell and aluminum sphere are concentric. (b) Find the electric field at all points in space, including points inside the aluminum and copper shell when the centres of copper shell and aluminum sphere are $1~\mathrm{cm}$ apart.

65. A long cylinder of aluminum of radius $R$ meters is charged so that it has a uniform charge per unit length on its surface of $\lambda$ . (a) Find the electric field inside and outside the cylinder. (b) Plot electric field as a function of distance from the centre of the rod.

66. At the surface of any conductor in electrostatic equilibrium, $E=\sigma/\epsilon_0$ . Show that this equation is consistent with the fact that $E=kq/r^2$ at the surface of a spherical conductor.

67. Two parallel plates $10~\mathrm{cm}$ on a side are given equal and opposite charges of magnitude $5.0\times10^{-9}~\mathrm{C}$ . The plates are $1.5~\mathrm{mm}$ apart. What is the electric field at the centre of the region between the plates?

68. Two parallel conducting plates, each of cross-sectional area $400~\mathrm{cm}^2$ , are $2.0~\mathrm{cm}$ apart and uncharged. If $1.0\times10^{12}$ electrons are transferred from one plate to the other, what are (a) the charge density on each plate? (b) The electric field between the plates?

69. The surface charge density on a long straight metallic pipe is $\sigma$ . What is the electric field outside and inside the pipe? Assume the pipe has a diameter of $2a$ .

Figure shows a pipe, with a cylindrical section highlighted. An arrow pointing up and one pointing down along the pipe from the cylinder are labeled infinity. There are plus signs inside the walls of the cylinder.

70. A point charge $q=-5.0\times10^{-12}~\mathrm{C}$ is placed at the centre of a spherical conducting shell of inner radius $3.5~\mathrm{cm}$ and outer radius $4.0~\mathrm{cm}$ . The electric field just above the surface of the conductor is directed radially outward and has magnitude $8.0~\mathrm{N/C}$ . (a) What is the charge density on the inner surface of the shell? (b) What is the charge density on the outer surface of the shell? (c) What is the net charge on the conductor?

71. A solid cylindrical conductor of radius $a$ is surrounded by a concentric cylindrical shell of inner radius $b$ . The solid cylinder and the shell carry charges $+Q$ and $-Q$ , respectively. Assuming that the length $L$ of both conductors is much greater than $a$ or $b$ , determine the electric field as a function of $r$ , the distance from the common central axis of the cylinders, for (a) $r<a$ ; (b) $a<r<b$ ; and (c) $r>b$ .

Additional Problems

72. A vector field $\vec{\mathbf{E}}$ (not necessarily an electric field; note units) is given by $\vec{\mathbf{E}}=3x^2\hat{\mathbf{k}}$ . Calculate $\int_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}da$ , where $S$ is the area shown below. Assume that $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ .
A square S with length of each side equal to a is shown in the xy plane.

A square S with length of each side equal to a is shown in the xy plane.

73. Repeat the preceding problem, with $\vec{\mathbf{E}}=2x\hat{\mathbf{i}}+3x^2\hat{\mathbf{k}}$ .

74. A circular area $S$ is concentric with the origin, has radius $a$ , and lies in the $yz$ -plane. Calculate $\int_S\vec{\mathbf{E}}\cdot\hat{\mathbf{n}}dA$ for $\vec{\mathbf{E}}=3z^2\hat{\mathbf{i}}$ .

75. (a) Calculate the electric flux through the open hemispherical surface due to the electric field $\vec{\mathbf{E}}=E_0\hat{\mathbf{k}}$ (see below). (b) If the hemisphere is rotated by $90^{\circ}$ around the $x$ -axis, what is the flux through it?

A hemisphere with radius R is shown with its base in the xy plane and center of base at the origin. An arrow is shown beside it, labeled vector E equal to E0 k hat.

76. Suppose that the electric field of an isolated point charge were proportional to $1/r^{2+\sigma}$ rather than $1/r^2$ . Determine the flux that passes through the surface of a sphere of radius $R$ centred at the charge. Would Gauss’s law remain valid?

77. The electric field in a region is given by $\vec{\mathbf{E}}=a/(b+cx)\hat{\mathbf{i}}$ where $a=200~\mathrm{N}\cdot\mathrm{m/C}$ , $b=2.00~\mathrm{m}$ and $c=2.0$ . What is the net charge enclosed by the shaded volume shown below?

Figure shows a cuboid with one corner on the origin of the coordinate axes. Its length along the x axis is 2 m, along y axis is 1.5 m and along z axis is 1 m. An arrow outside the cuboid points along the x axis. It is labeled vector E.

78. Two equal and opposite charges of magnitude $Q$ are located on the x-axis at the points $+a$ and $-a$ , as shown below. What is the net flux due to these charges through a square surface of side $2a$ that lies in the $yz$ -plane and is centred at the origin? (Hint:Determine the flux due to each charge separately, then use the principle of superposition. You may be able to make a symmetry argument.)
A shaded square is shown in the yz plane with its center at the origin. Its side parallel to z axis is labeled to be of length 2a. A charge labeled plus Q is shown on the positive x axis at a distance a from the origin. A charge labeled minus Q is shown on the negative x axis at a distance a from the origin.

79. A fellow student calculated the flux through the square for the system in the preceding problem and got $0$ . What went wrong?

80. A $10~\mathrm{cm}\times10~\mathrm{cm}$ piece of aluminum foil of $0.1~\mathrm{mm}$ thickness has a charge of $20~\mu\mathrm{C}$ that spreads on both wide side surfaces evenly. You may ignore the charges on the thin sides of the edges. (a) Find the charge density. (b) Find the electric field $1~\mathrm{cm}$ from the centre, assuming approximate planar symmetry.

81. Two $10~\mathrm{cm}\times10~\mathrm{cm}$ pieces of aluminum foil of thickness $0.1~\mathrm{mm}$ face each other with a separation of $5~\mathrm{mm}$ . One of the foils has a charge of $+30~\mu\mathrm{C}$ and the other has $-30~\mu\mathrm{C}$ . (a) Find the charge density at all surfaces, i.e., on those facing each other and those facing away. (b) Find the electric field between the plates near the center assuming planar symmetry.

82. Two large copper plates facing each other have charge densities $\pm4.0~\mathrm{C/m}^2$ on the surface facing the other plate, and zero in between the plates. Find the electric flux through a $3~\mathrm{cm}\times4~\mathrm{cm}$ rectangular area between the plates, as shown below, for the following orientations of the area. (a) If the area is parallel to the plates, and (b) if the area is tilted $\theta=30^{\circ}$ from the parallel direction. Note, this angle can also be $\theta=180^{\circ}+30^{\circ}$ .

Figure shows two parallel plates and a dotted line exactly between the two, parallel to them. A third plate forms an angle theta with the dotted line.

83. The infinite slab between the planes defined by $z=-a/2$ and $z=a/2$ contains a uniform volume charge density $\rho$ (see below). What is the electric field produced by this charge distribution, both inside and outside the distribution?

84. A total charge $Q$ is distributed uniformly throughout a spherical volume that is centred at $O_1$ and has a radius $R$ . Without disturbing the charge remaining, charge is removed from the spherical volume that is centred at $O_2$ (see below). Show that the electric field everywhere in the empty region is given by $\vec{\mathbf{E}}=\frac{Q\vec{\mathbf{r}}}{4\pi\epsilon_0R^3}$ where $\vec{\mathbf{r}}$ is the displacement vector directed from $O_1$ to $O_2$ .

Figure shows a circle with center O1 and radius R. Another smaller circle with center O2 is shown within it. An arrow from O1 to O2 is labeled vector r.

85. A non-conducting spherical shell of inner radius $a_1$ and outer radius

Introduction to Electricity, Magnetism, and Circuits

Introduction to Electricity, Magnetism, and Circuits

Daryl Janzen

University of Saskatchewan, Distance Education Unit

Saskatoon, SK, Canada

Contents

1

Introduction

2

Acknowledgements

1

1 Electric Charges and Fields

Chapter Outline

2

1.1 Electric Charge

LEARNING OBJECTIVES

Discoveries

Properties of Electric Charge

The Source of Charges: The Structure of the Atom

A Note on Terminology

3

1.2 Conductors, Insulators, and Charging by Induction

LEARNING OBJECTIVES

Conductors and Insulators

Charging by Induction

4

1.3 Coulomb's Law

LEARNING OBJECTIVES

COULOMB’S LAW

EXAMPLE 1.3.1

The Force on the Electron in Hydrogen

Strategy

Solution

Significance

CHECK YOUR UNDERSTANDING 1.1

Multiple Source Charges

EXAMPLE 1.3.2

The Net Force from Two Source Charges

Strategy

Solution

Significance

CHECK YOUR UNDERSTANDING 1.2

5

1.4 Electric Field

LEARNING OBJECTIVES

Defining a Field

The Meaning of “Field”

Superposition

The Direction of the Field

DIRECTION OF THE ELECTRIC FIELD

INTERACTIVE

EXAMPLE 1.4.1

The E-field of an Atom

Strategy

Solution

EXAMPLE 1.4.2

The E-Field above Two Equal Charges

Strategy

Solution

Significance

CHECK YOUR UNDERSTANDING 1.3

INTERACTIVE

6

1.5 Calculating Electric Fields of Charge Distributions

LEARNING OBJECTIVES

EXAMPLE 1.5.1

Electric Field of a Line Segment

Strategy

Solution

Significance

CHECK YOUR UNDERSTANDING 1.4

EXAMPLE 1.5.2

Electric Field of an Infinite Line of Charge

Strategy

Solution

Significance

EXAMPLE 1.5.3

Electric Field due to a Ring of Charge

Strategy

Solution