Introduction to Electricity, Magnetism, and Circuits by Daryl Janzen is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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Cover image: “The Auroral Radar” by Ashton Reimer won first prize in the “Research in Action” category in the University of Saskatchewan 2016 Images of Research contest, and is used here with permission. It depicts the aurora borealis, or Northern Lights, seen over the Saskatoon SuperDARN (Super Dual Auroral Radar Network) radar. On December 20th, 2015, a large geomagnetic storm produced this show, which was caused by the impact of two successive coronal mass ejections from the Sun. While storms produce beautiful aurora, they also produce adverse effects on airplane communications systems, GPS, and the electrical power grid. SuperDARN radars measure the velocity of the aurora, in a manner similar to a police radar gun, and this radar data is an essential tool used in space weather forecasting, which can predict the intensity of these effects.
Introduction to Electricity, Magnetism, and Circuits by Daryl Janzen is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted
1.1 Electric Charge
1.2 Conductors, Insulators, and Charging by Induction
1.3 Coulomb’s Law
1.4 Electric Field
1.5 Calculating Electric Fields of Charge Distributions
1.6 Electric Field Lines
1.7 Electric Dipoles
Chapter 1 Review
Normally it is through the study of Newton’s laws, which govern the motions of everyday objects, that we first introduce students to the mathematical concept of force. Several physical phenomena can be identified as forces based on the effect they have on a physical object: Specifically, they cause objects to accelerate, to change their momentum. Thus, a force is recognised by the effect that it has on an object.
Gravitation, for example, is a phenomenon that is identified as a force that acts on all objects with mass. In this chapter, we begin the study of the electric force, which acts on all objects with a property called charge. The electric force is much stronger than gravity (in most systems where both appear), but it can be a force of attraction or a force of repulsion, which leads to very different effects on objects. The electric force helps bind atoms together, so it is of fundamental importance in matter. But it also governs most everyday interactions we deal with, from chemical interactions to biological processes.
By the end of this section, you will be able to:
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.
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)
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)
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)
(Figure 1.1.4)
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 semiprecious 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 socalled “longrange” 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)
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:
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.
In addition to the existence of two types of charge, several other properties of charge have been discovered.
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 chargetomass 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 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)
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 lowmass 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)
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.
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 .” However, it is very common to say instead, “Suppose we have a – 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.
By the end of this section, you will be able to:
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)
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 semiprecious gems are insulators, as are materials like wood, glass, and plastic.
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)
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)
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)
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)
By the end of this section, you will be able to:
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
The electric force 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:
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.)
The electric force (or Coulomb force) between two electrically charged particles is equal to
(1.3.1)
We use absolute value signs around the product because one of the charges may be negative, but the magnitude of the force is always positive. The unit vector points directly from the charge toward . If and have the same sign, the force vector on points away from ; if they have opposite signs, the force on points toward (Figure 1.3.1).
(Figure 1.3.1)
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 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 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
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:
A hydrogen atom consists of a single proton and a single electron. The proton has a charge of and the electron has . In the “ground state” of the atom, the electron orbits the proton at most probable distance of (Figure 1.3.2). Calculate the electric force on the electron due to the proton.
(Figure 1.3.2)
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.
Our two charges and the distance between them are,
The magnitude of the force on the electron is
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
This is a threedimensional 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.
What would be different if the electron also had a positive charge?
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 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 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
In this expression, represents the charge of the particle that is experiencing the electric force , and is located at from the origin; the ’s are the source charges, and the vectors are the displacements from the position of the th charge to the position of . Each of the 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 and the difference in labels is merely to allow clear discussion, with being the charge we are determining the force on.
(Figure 1.3.3)
(Note that the force vector does not necessarily point in the same direction as the unit vector ; it may point in the opposite direction, . 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.
Three different, small charged objects are placed as shown in Figure 1.3.4. The charges and are fixed in place; is free to move. Given , , and , and that what is the net force on the middle charge ?
(Figure 1.3.4)
We use Coulomb’s law again. The way the question is phrased indicates that is our test charge, so that and are source charges. The principle of superposition says that the force on from each of the other charges is unaffected by the presence of the other charge. Therefore, we write down the force on from each and add them together as vectors.
We have two source charges ( and ), a test charge (), distances ( and ), and we are asked to find a force. This calls for Coulomb’s law and superposition of forces. There are two forces:
We can’t add these forces directly because they don’t point in the same direction: points only in the direction, while points only in the direction. The net force is obtained from applying the Pythagorean theorem to its and components:
where
and
We find that
at an angle of
that is, above the axis, as shown in the diagram.
Notice that when we substituted the numerical values of the charges, we did not include the negative sign of either or 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.
What would be different if were negative?
1.3 Coulomb's Law by Daryl Janzen is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
By the end of this section, you will be able to:
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.
Suppose we have source charges located at positions , applying electrostatic forces on a test charge . The net force on is (see Equation 1.3.2)
We can rewrite this as
where
or, more compactly,
This expression is called the electric field at position of the source charges. Here, is the location of the point in space where you are calculating the field and is relative to the positions 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)
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 of a point charge is similar to the gravitational field 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 , and then we calculate the resulting force (i.e., weight) on different masses. Also, the general expression for calculating at arbitrary distances from the center of Earth (i.e., not just near Earth’s surface) is very similar to the expression for : , where is a proportionality constant, playing the same role for as does for. The value of 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 , the units of are newtons per coulomb, , that is, the electric field applies a force on each unit charge. Now notice the units of : From , the units of are newtons per kilogram, , 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 .
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 (both the magnitude and the direction) depends on where in space the point is located, measured from the locations of the source charges .
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 chargebearing object (the source charge) as a continuous, immaterial substance that surrounds the source charge, filling all of space—in principle, to 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.
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 at position , then calculate the field of at , while—and this is the crucial idea—ignoring the field of, and indeed even the existence of, . 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.
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).
By convention, all electric fields point away from positive source charges and point toward negative source charges.
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.
In an ionized helium atom, the most probable distance between the nucleus and the electron is . What is the electric field due to the nucleus at the location of the electron?
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 (Figure 1.4.2).
(Figure 1.4.2)
The electric field is calculated by
Since there is only one source charge (the nucleus), this expression simplifies to
Here (since there are two protons) and is given; substituting gives
The direction of 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.
(a) Find the electric field (magnitude and direction) a distance above the midpoint between two equal charges that are a distance apart (Figure 1.4.3). Check that your result is consistent with what you’d expect when .
(b) The same as part (a), only this time make the righthand charge instead of .
(Figure 1.4.3)
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: , and constants .
(a) By symmetry, the horizontal ()components of cancel (Figure 1.4.4);
(Figure 1.4.4)
The vertical ()component is given by
What we want to do now is replace the quantities in this expression that we don’t know (such as ), or can’t easily measure (such as ) with quantities that we do know, or can measure. In this case, by geometry,
and
Thus, substituting,
Simplifying, the desired answer is
(1.4.3)
(b) If the source charges are equal and opposite, the vertical components cancel because
and we get, for the horizontal component of,
This becomes
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 , , and , 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., ), the two source charges should “merge” and we should then “see” the field of just one charge, of size . So, let ; then we can neglect in Equation 1.4.3 to obtain
which is the correct expression for a field at a distance away from a charge .
Next, we consider the field of equal and opposite charges, Equation 1.4.4. It can be shown (via a Taylor expansion) that for , this becomes
which is the field of a dipole, a system that we will study in more detail later. (Note that the units of are still correct in this expression, since the units of in the numerator cancel the unit of the “extra” in the denominator.) If is very large , then , as it should; the two charges “merge” and so cancel out.
What is the electric field due to a single point particle?
Try this simulation of electric field hockey to get the charge in the goal by placing other charges on the field.
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 H2OH2O 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)
Definitions of charge density:
Then, for a line charge, a surface charge, and a volume charge, the summation in Equation 1.4.2 becomes an integral and is replaced by , , or respectively:
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 , , or as the case may be, expressed in terms of , and also expressing the charge density function appropriately. It may be constant; it might be dependent on location.
Note carefully the meaning of in these equations: It is the distance from the charge element to the location of interest, (the point in space where you want to determine the field). However, don’t confuse this with the meaning of ; we are using it and the vector notation to write three integrals at once. That is, Equation 1.5.2 is actually
Find the electric field a distance above the midpoint of a straight line segment of length that carries a uniform line charge density .
Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length , each of which carries a differential amount of charge . 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)
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
The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal ()components of the field cancel, so that the net field points in the direction. Let’s check this formally.
The total field is the vector sum of the fields from each of the two charge elements (call them and , for now):
Because the two charge elements are identical and are the same distance away from the point where we want to calculate the field, , so those components cancel. This leaves
These components are also equal, so we have
where our differential line element is , in this example, since we are integrating along a line of charge that lies on the axis. (The limits of integration are to , not to , because we have constructed the net field from two differential pieces of charge . If we integrated along the entire length, we would pick up an erroneous factor of .)
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 and 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
and
Substituting, we obtain
which simplifies to
(1.5.5)
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.
How would the strategy used above change to calculate the electric field at a point a distance above one end of the finite line segment?
Find the electric field a distance above the midpoint of an infinite line of charge that carries a uniform line charge density .
This is exactly like the preceding example, except the limits of integration will be to .
Again, the horizontal components cancel out, so we wind up with
where our differential line element is , in this example, since we are integrating along a line of charge that lies on the axis. Again,
Substituting, we obtain
which simplifies to
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 , dominates the in the denominator, so that Equation 1.5.5 simplifies to
If you recall that , the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.
In the limit , 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:
An interesting artifact of this infinite limit is that we have lost the usual dependence that we are used to. This will become even more intriguing in the case of an infinite plane.
A ring has a uniform charge density , 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.
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)
The electric field for a line charge is given by the general expression
A general element of the arc between and is of length and therefore contains a charge equal to . The element is at a distance of from , the angle is , and therefore the electric field is
As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when we take the limit of , we find that
as we expect.
Find the electric field of a circular thin disk of radius and uniform charge density at a distance above the centre of the disk (Figure 1.5.4).
(Figure 1.5.4)
The electric field for a surface charge is given by
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 direction. The vertical component of the electric field is extracted by multiplying by , so
As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities. In this case,
(Please take note of the two different “s” here; is the distance from the differential ring of charge to the point where we wish to determine the field, whereas 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 .
Substituting all this in, we get
or, more simply,
(1.5.7)
Again, it can be shown (via a Taylor expansion) that when , this reduces to
which is the expression for a point charge .
How would the above limit change with a uniformly charged rectangle instead of a disk?
As , 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:
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 direction.
Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities (Figure 1.5.5).
(Figure 1.5.5)
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.
The electric field points away from the positively charged plane and toward the negatively charged plane. Since the 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
for the electric field. The is because in the figure, the field is pointing in the direction.
Systems that may be approximated as two infinite planes of this sort provide a useful means of creating uniform electric fields.
What would the electric field look like in a system with two parallel positively charged planes with equal charge densities?
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 threedimensional. Although in this book we include some pseudothreedimensional images, several of the diagrams that you’ll see (both here, and in subsequent chapters) will be twodimensional projections, or crosssections. Always keep in mind that in fact, you’re looking at a threedimensional 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)
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 threedimensional; 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)
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)
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 crosssectional area perpendicular to the electric field. This field line density is drawn to be proportional to the magnitude of the field at that crosssection. 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 crosssection, this indicates the magnitude of the field is small. Figure 1.6.4 shows the idea.
(Figure 1.6.4)
In Figure 1.6.4, the same number of field lines passes through both surfaces ( and ), but the surface is larger than surface . Therefore, the density of field lines (number of lines per unit area) is larger at the location of , indicating that the electric field is stronger at the location of than at . The rules for creating an electric field diagram are as follows.
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 charge as entering the . The field lines that do not terminate at emanate outward from the charge configuration, to infinity.
(Figure 1.6.5)
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 computerdrawn diagrams.
One example of a fieldline drawing program is from the PhET “Charges and Fields” simulation.
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 between the two charges is much, much less than the distance of the field point , 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 . We assume that the dipole is a permanent dipole; it exists without the field, and does not break apart in the external 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 , 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:
The quantity (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 socalled dipole moment of the dipole:
We can therefore write
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 ) so that it is parallel to the direction of the external field.
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 “builtin” 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)
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 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)
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:
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.
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 onedimensional (the width and height are much, much smaller than its length); its units are
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 twodimensional charge distribution (the thickness is small); its units are
volume charge density
amount of charge in an element of a threedimensional charge distribution; its units are
Coulomb’s law 

Superposition of electric forces 

Electric force due to an electric field  
Electric field at point 

Field of an infinite wire 

Field of an infinite plane 

Dipole moment 

Torque on dipole in external field 

where and are two point charges separated by a distance . This Coulomb force is extremely basic, since most charges are due to pointlike particles. It is responsible for all electrostatic effects and underlies most macroscopic forces.
1.1 The force would point outward.
1.2 The net force would point below the axis.
1.3
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 where and are the sides of the rectangle but otherwise identical.
1.6 The electric field would be zero in between, and have magnitude everywhere else.
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?
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?
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 near a large metal plate. (a) If is attracted to the plate, is the plate necessarily charged? (b) If is repelled by the plate, is the plate necessarily charged?
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?
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 axis. Is it true that the net electric field always vanishes at some point (other than infinity) along the axis?
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?
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 and a charge ?
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?
37. Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of ? (b) How many electrons must be removed from a neutral object to leave a net charge of ?
38. If 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 electrons through the starter motor. How many coulombs of charge were moved?
40. A certain lightning bolt moves of charge. How many fundamental units of charge is this?
41. A – copper penny is given a charge of . (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 – copper penny is given a charge of . (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?
43. Suppose a speck of dust in an electrostatic precipitator has protons in it and has a net charge of (a very large charge for a small speck). How many electrons does it have?
44. An amoeba has protons and a net charge of . (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 – ball of copper has a net charge of . What fraction of the copper’s electrons has been removed? (Each copper atom has protons, and copper has an atomic mass of .)
46. What net charge would you place on a – piece of sulfur if you put an extra electron on in of its atoms? (Sulfur has an atomic mass of .)
47. How many coulombs of positive charge are there in of plutonium, given its atomic mass is and that each plutonium atom has protons?
48. Two point particles with charges and are held in place by – forces on each charge in appropriate directions. (a) Draw a freebody diagram for each particle. (b) Find the distance between the charges.
49. Two charges and are fixed apart, with the second one to the right. Find the magnitude and direction of the net force on a – charge when placed at the following locations: (a) halfway between the two (b) half a meter to the left of the charge (c) half a meter above the 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 . What is the force of attraction between the two singly charged ions?
51. Protons in an atomic nucleus are typically apart. What is the electric force of repulsion between nuclear protons?
52. Suppose Earth and the Moon each carried a net negative charge . Approximate both bodies as point masses and point charges. (a) What value of 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 and are placed apart. What is the force on a third charge placed midway between and ?
54. Where must of the preceding problem be placed so that the net force on it is zero?
55. Two small balls, each of mass , are attached to silk threads long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge , the threads hang at to the vertical, as shown below. What is the magnitude of ? What are the signs of the two charges?
56. Point charges and are located at and . What is the force of on ?
57. The net excess charge on two small spheres (small enough to be treated as point charges) is . Show that the force of repulsion between the spheres is greatest when each sphere has an excess charge . 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 when they are apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of . What is the original charge on each sphere?
59. A charge is placed at the point shown below. What is the force on ?
60. What is the net electric force on the charge located at the lower righthand corner of the triangle shown here?
61. Two fixed particles, each of charge , are apart. What force do they exert on a third particle of charge that is from each of them?
62. The charges , , and are placed at the corners of the triangle shown below. What is the force on ?
63. What is the force on the charge at the lowerrighthand corner of the square shown here?
64. Point charges and are fixed at and. What is the force of on ?
65. A particle of charge experiences an upward force of magnitude 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 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 . Compare the gravitational and electric forces on a small dust particle of mass that carries a single electron charge. What is the acceleration (both magnitude and direction) of the dust particle?
67. Consider an electron that is from an alpha particle (). (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 and has a mass . The length of each thread is , and at equilibrium, the balls are separated by an angle . How does vary with and ? Show that satisfies .
69. What is the electric field at a point where the force on a charge is ?
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 . What is the acceleration of an electron in this field?
72. A small piece of cork whose mass is is given a charge of . What electric field is needed to place the cork in equilibrium under the combined electric and gravitational forces?
73. If the electric field is at a distance of from a point charge , what is the value of ?
74. What is the electric field of a proton at the first Bohr orbit for hydrogen ()? 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 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, and , are held apart. (a) What is the electric field at a point 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 and are placed apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge situated there?
78. Can you arrange the two point charges and along the axis so that at the origin?
79. Point charges are fixed on the axis at and . What charge must be placed at the origin so that the electric field vanishes at ,
80. A thin conducting plate on the side is given a charge of . An electron is placed 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 .
82. Two thin conducting plates, each on a side, are situated parallel to one another and apart. If 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 . What is the electric field at the point ? (Hint: Solve this problem by first considering the electric field at due to a small segment of the rod, which contains charge . Then find the net field by integrating over the length of the rod.)
84. The charge per unit length on the thin semicircular wire shown below is . What is the electric field at the point ?
85. Two thin parallel conducting plates are placed apart. Each plate is on a side; one plate carries a net charge of and the other plate carries a net charge of 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 on a side is given a total charge of (a) What is the electric field above the plate? (b) What is the force on an electron at this point? (c) Repeat these calculations for a point above the plate. (d) When the electron moves from to above the plate, how much work is done on it by the electric field?
87. A total charge is distributed uniformly along a thin, straight rod of length (see below). What is the electric field at ? At ?
88. Charge is distributed along the entire axis with uniform density . How much work does the electric field of this charge distribution do on an electron that moves along the axis from to ?
89. Charge is distributed along the entire axis with uniform density and along the entire axis with uniform density . Calculate the resulting electric field at (a) and (b)
90. A rod bent into the arc of a circle subtends an angle at the centre of the circle (see below). If the rod is charged uniformly with a total charge , what is the electric field at ?
91. A proton moves in the electric field . (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 . Determine the distance and time for each particle to acquire a kinetic energy of .
93. A spherical water droplet of radius carries an excess 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 and the speed of the proton when it enters is What distance has the proton been deflected downward when it leaves the plates?
95. Shown below is a small sphere of mass that carries a charge of The sphere is attached to one end of a very thin silk string long. The other end of the string is attached to a large vertical conducting plate that has a charge density of What is the angle that the string makes with the vertical?
96. Two infinite rods, each carrying a uniform charge density are parallel to one another and perpendicular to the plane of the page. (See below.) What is the electrical field at ? At ?
97. Positive charge is distributed with a uniform density along the positive axis from to , along the positive axis from to , and along a arc of a circle of radius as shown below. What is the electric field at ?
98. From a distance of , a proton is projected with a speed of directly at a large, positively charged plate whose charge density is . (See below.) (a) Does the proton reach the plate? (b) If not, how far from the plate does it turn around?
99. A particle of mass and charge moves along a straight line away from a fixed particle of charge . When the distance between the two particles is is moving with a speed . (a) Use the workenergy theorem to calculate the maximum separation of the charges. (b) What do you have to assume about v0v0 to make this calculation? (c) What is the minimum value of such that escapes from ?
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 and situated from each other. (b) Draw the electric field lines map for two charges and situated from each other. (c) Draw the electric field lines map for two charges and situated from each other.
102. Draw the electric field for a system of three particles of charges and fixed at the corners of an equilateral triangle of side .
103. Two charges of equal magnitude but opposite sign make up an electric dipole. A quadrupole consists of two electric dipoles are placed antiparallel at two edges of a square as shown.
Draw the electric field of the charge distribution.
104. Suppose the electric field of an isolated point charge decreased with distance as rather than as . Show that it is then impossible to draw continuous field lines so that their number per unit area is proportional to .
106. (a) What is the dipole moment of the configuration shown above? If , (b) what is the torque on this dipole with an electric field of ? (c) What is the torque on this dipole with an electric field of? (d) What is the torque on this dipole with an electric field of ?
108. Point charges and are located at and . What is the force of on ?
109. What is the force on the – charge shown below?
110. What is the force on the – charge placed at the centre of the square shown below?
111. Four charged particles are positioned at the corners of a parallelogram as shown below. If and what is the net force on ?
112. A charge is fixed at the origin and a second charge moves along the axis, as shown below. How much work is done on by the electric force when moves from to
113. A charge is released from rest when it is from a fixed charge What is the kinetic energy of when it is from ?
114. What is the electric field at the midpoint of the hypotenuse of the triangle shown below?
115. Find the electric field at for the charge configurations shown below.
116. (a) What is the electric field at the lowerrighthand corner of the square shown below? (b) What is the force on a charge placed at that point?
117. Point charges are placed at the four corners of a rectangle as shown below: and What is the electric field at ?
118. Three charges are positioned at the corners of a parallelogram as shown below. (a) If what is the electric field at the unoccupied corner? (b) What is the force on a – charge placed at this corner?
119. A positive charge is released from rest at the origin of a rectangular coordinate system and moves under the influence of the electric field What is the kinetic energy of when it passes through ?
120. A particle of charge and mass is placed at the centre of a uniformly charged ring of total charge and radius . 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
121. Charge is distributed uniformly along the entire axis with a density and along the positive axis from to with a density . What is the force between the two distributions?
122. The circular arc shown below carries a charge per unit length where is measured from the axis. What is the electric field at the origin?
123. Calculate the electric field due to a uniformly charged rod of length , aligned with the axis with one end at the origin; at a point on the axis.
124. The charge per unit length on the thin rod shown below is What is the electric force on the point charge ? Solve this problem by first considering the electric force on due to a small segment of the rod, which contains charge Then, find the net force by integrating over the length of the rod.
125. The charge per unit length on the thin rod shown here is What is the electric force on the point charge ? (See the preceding problem.)
126. The charge per unit length on the thin semicircular wire shown below is What is the electric force on the point charge ? (See the preceding problems.)
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
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)
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)
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 of area that is perpendicular to the uniform electric field . If field lines pass through , then we know from the definition of electric field lines (Electric Charges and Fields) that , or .
The quantity is the electric flux through . We represent the electric flux through an open surface like by the symbol . Electric flux is a scalar quantity and has an SI unit of newtonmeters squared per coulomb (). Notice that may also be written as , demonstrating that electric flux is a measure of the number of field lines crossing a surface.
(Figure 2.1.2)
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 of area that is inclined at an angle to the plane and whose projection in that plane is (area ). The areas are related by . Because the same number of field lines crosses both and , the fluxes through both surfaces must be the same. The flux through is therefore . Designating as a unit vector normal to (see Figure 2.1.2(b)), we obtain
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.
For discussing the flux of a vector field, it is helpful to introduce an area vector . 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 has the following magnitude and direction:
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)
Since 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)
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), is chosen to be the outward normal at every point, to be consistent with the sign convention for electric charge.
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:
Figure 2.1.5 shows the electric field of an oppositely charged, parallelplate 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)
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, is positive; when they enter (or “flow into”) the surface, is negative.
Any smooth, nonflat 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)
To keep track of the patches, we can number them from 1 through . 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 th patch by . (We have used the symbol 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 th patch by .
Therefore, we can write the electric flux through the area of the th patch as
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 , which we denote simply as .
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 and unit normal . Since the elements are infinitesimal, they may be assumed to be planar, and may be taken as constant over any element. Then the flux through an area is given by . It is positive when the angle between and is less than and negative when the angle is greater than . 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 representing the integral over ,
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
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.
A constant electric field of magnitude points in the direction of the positive axis (Figure 2.1.7). What is the electric flux through a rectangle with sides and in the (a) plane and in the (b) plane?
(Figure 2.1.7)
Apply the definition of flux: , where the definition of dot product is crucial.
The relative directions of the electric field and area can cause the flux through the area to be zero.
A constant electric field of magnitude points in the direction of the positive axis (Figure 2.1.8). What is the net electric flux through a cube?
(Figure 2.1.8)
Apply the definition of flux: , noting that a closed surface eliminates the ambiguity in the direction of the area vector.
Through the top face of the cube,
Through the bottom face of the cube, 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 .
The net flux of a uniform electric field through a closed surface is zero.
A uniform electric field of magnitude is directed parallel to the plane at above the plane, as shown in Figure 2.1.9. What is the electric flux through the plane surface of area located in the plane? Assume that points in the positive direction.
(Figure 2.1.9)
Apply , where the direction and magnitude of the electric field are constant.
The angle between the uniform electric field and the unit normal to the planar surface is . Since both the direction and magnitude are constant, comes outside the integral. All that is left is a surface integral over , which is . Therefore, using the opensurface equation, we find that the electric flux through the surface is
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.
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?
What is the total flux of the electric field through the rectangular surface shown in Figure 2.1.10?
(Figure 2.1.10)
Apply . We assume that the unit normal to the given surface points in the positive direction, so . Since the electric field is not uniform over the surface, it is necessary to divide the surface into infinitesimal strips along which is essentially constant. As shown in Figure 2.1.10, these strips are parallel to the xaxis, and each strip has an area .
From the open surface integral, we find that the net flux through the rectangular surface is
For a nonconstant electric field, the integral method is required.
If the electric field in Example 2.1.4 is , what is the flux through the rectangular area?
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
where is the radial vector from the charge at the origin to the point . We can use this electric field to find the flux through the spherical surface of radius , as shown in Figure 2.2.1.
(Figure 2.2.1)
Then we apply to this system and substitute known values. On the sphere, and , so for an infinitesimal area ,
We now find the net flux by integrating this flux over the surface of the sphere:
where the total surface area of the spherical surface is . This gives the flux through the closed spherical surface at radius as
(2.2.1)
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 with distance, which just cancels the rate of increase of the surface area.
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 that pierces the surface at radius also pierces the surface at (Figure 2.2.2).
(Figure 2.2.2)
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 and leaves at . 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)
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 through any closed surface, also called a Gaussian surface, is equal to the net charge enclosed () divided by the permittivity of free space ():
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 or is negative.
(Figure 2.2.4)
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 of Figure 2.2.5 is . Note that 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)
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.
To use Gauss’s law effectively, you must have a clear understanding of what each term in the equation represents. The field 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, 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)
Calculate the electric flux through each Gaussian surface shown in Figure 2.2.7.
(Figure 2.2.7)
From Gauss’s law, the flux through each surface is given by , where is the charge enclosed by that surface.
For the surfaces and charges shown, we find
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.
Calculate the electric flux through the closed cubical surface for each charge distribution shown in Figure 2.2.8.
(Figure 2.2.8)
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.
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 over which the electric field has constant magnitude. Furthermore, if is parallel to everywhere on the surface, then . (If and are antiparallel everywhere on the surface, then.) Gauss’s law then simplifies to
where 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
The direction of the electric field at the field point 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 . Here is a summary of the steps we will follow:
Basically, there are only three types of symmetry that allow Gauss’s law to be used to deduce the electric field. They are
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.
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 is uniformly charged with charge density then the distribution has spherical symmetry (Figure 2.3.1(a)). On the other hand, if a sphere of radius is charged so that the top half of the sphere has uniform charge density and the bottom half has a uniform charge density , 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)
One good way to determine whether or not your problem has spherical symmetry is to look at the charge density function in spherical coordinates, . If the charge density is only a function of , that is , then you have spherical symmetry. If the density depends on or you could change it by rotation; hence, you would not have spherical 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 located at a distance from the centre:
where is the unit vector pointed in the direction from the origin to the field point . The radial component of the electric field can be positive or negative. When , the electric field at points away from the origin, and when , the electric field at points toward the origin.
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)
The magnitude of the electric field must be the same everywhere on a spherical Gaussian surface concentric with the distribution. For a spherical surface of radius ,
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 . Let be the total charge enclosed inside the distance from the origin, which is the space inside the Gaussian spherical surface of radius . This gives the following relation for Gauss’s law:
Hence, the electric field at point that is a distance from the centre of a spherically symmetrical charge distribution has the following magnitude and direction:
Direction: radial from to or from to .
The direction of the field at point 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 to , and for a net negative charge, the direction is from to . 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.
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 of the field point relative to the radius of the charge distribution , such as that shown in Figure 2.3.3.
(Figure 2.3.3)
If point is located outside the charge distribution—that is, if —then the Gaussian surface containing encloses all charges in the sphere. In this case, equals the total charge in the sphere. On the other hand, if point is within the spherical charge distribution, that is, if , then the Gaussian surface encloses a smaller sphere than the sphere of charge distribution. In this case, is less than the total charge present in the sphere. Referring to Figure 2.3.3, we can write as
The field at a point outside the charge distribution is also called , and the field at a point inside the charge distribution is called. 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)
(2.3.5)
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.
A sphere of radius , such as that shown in Figure 2.3.3, has a uniform volume charge density . Find the electric field at a point outside the sphere and at a point inside the sphere.
Apply the Gauss’s law problemsolving strategy, where we have already worked out the flux calculation.
The charge enclosed by the Gaussian surface is given by
The answer for electric field amplitude can then be written down immediately for a point outside the sphere, labeled , and a point inside the sphere, labeled .
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 , whereas the field from each infinitesimal element of charge drops off 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)
The direction of the electric field at any point is radially outward from the origin if is positive, and inward (i.e., toward the centre) if is negative. The electric field at some representative space points are displayed in Figure 2.3.5 whose radial coordinates are , , and .
(Figure 2.3.5)
Notice that 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.
A nonconducting sphere of radius has a nonuniform charge density that varies with the distance from its centre as given by
where is a constant. We require so that the charge density is not undefined at . Find the electric field at a point outside the sphere and at a point inside the sphere.
Apply the Gauss’s law strategy given above, where we work out the enclosed charge integrals separately for cases inside and outside the sphere.
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 , which depends on the location of the field point.
A note about symbols: We use for locating charges in the charge distribution and for locating the field point(s) at the Gaussian surface(s). The letter 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 and , 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 and the thickness . Multiplying the volume with the density at this location, which is , gives the charge in the shell:
(a) Field at a point outside the charge distribution. In this case, the Gaussian surface, which contains the field point , has a radius that is greater than the radius of the charge distribution, . Therefore, all charges of the charge distribution are enclosed within the Gaussian surface. Note that the space between and is empty of charges and therefore does not contribute to the integral over the volume enclosed by the Gaussian surface:
This is used in the general result for above to obtain the electric field at a point outside the charge distribution as
where 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 . Therefore, only those charges in the distribution that are within a distance of the centre of the spherical charge distribution count in :
Now, using the general result above for we find the electric field at a point that is a distance from the centre and lies within the charge distribution as
where the direction information is included by using the unit radial vector.
Check that the electric fields for the sphere reduce to the correct values for a point charge.
A charge distribution has cylindrical symmetry if the charge density depends only upon the distance 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 . in an infinite straight wire has a cylindrical symmetry, and so does an infinitely long cylinder with constant charge density . An infinitely long cylinder that has different charge densities along its length, such as a charge density for and for , 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 for and for . 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)
In all cylindrically symmetrical cases, the electric field at any point must also display cylindrical symmetry.
Cylindrical symmetry: ,
where is the distance from the axis and is a unit vector directed perpendicularly away from the axis (Figure 2.3.8).
(Figure 2.3.8)
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 and height 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)
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
whereas the flux through the end caps is zero because there. Thus, the flux is
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 , you find that of Gauss’s law is directly proportional to . Let us write it as charge per unit length () times length :
Hence, Gauss’s law for any cylindrically symmetrical charge distribution yields the following magnitude of the electric field a distance away from the axis:
The charge per unit length depends on whether the field point is inside or outside the cylinder of charge distribution, just as we have seen for the spherical distribution.
Let be the radius of the cylinder within which charges are distributed in a cylindrically symmetrical way. Let the field point be at a distance s from the axis. (The side of the Gaussian surface includes the field point .) When (that is, when is outside the charge distribution), the Gaussian surface includes all the charge in the cylinder of radius and length . When ( is located inside the charge distribution), then only the charge within a cylinder of radius and length is enclosed by the Gaussian surface:
A very long nonconducting cylindrical shell of radius has a uniform surface charge density . Find the electric field (a) at a point outside the shell and (b) at a point inside the shell.
Apply the Gauss’s law strategy given earlier, where we treat the cases inside and outside the shell separately.
(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 and length , 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 . Therefore, is given by
(Figure 2.3.10)
Hence, the electric field at a point outside the shell at a distance away from the axis is
where is a unit vector, perpendicular to the axis and pointing away from it, as shown in the figure. The electric field at points in the direction of given in Figure 2.3.10 if and in the opposite direction to if .
(b) Electric field at a point inside the shell. For a point inside the cylindrical shell, the Gaussian surface is a cylinder whose radius is less than (Figure 2.3.11). This means no charges are included inside the Gaussian surface:
(Figure 2.3.11)
This gives the following equation for the magnitude of the electric field at a point whose is less than of the shell of charges.
This gives us
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 .
A thin straight wire has a uniform linear charge density . Find the electric field at a distance from the wire, where is much less than the length of the wire.
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.
We take the plane of the charge distribution to be the plane and we find the electric field at a space point with coordinates . Since the charge density is the same at all coordinates in the plane, by symmetry, the electric field at cannot depend on the – or coordinates of point , as shown in Figure 2.3.12. Therefore, the electric field at 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 has only a nonzero component.
Uniform charges in plane:
where is the distance from the plane and is the unit vector normal to the plane. Note that in this system, , although of course they point in opposite directions.
(Figure 2.3.12)
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 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)
Let be the area of the shaded surface on each side of the plane and be the magnitude of the electric field at point . 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: .
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
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 . From Figure 2.3.13, we see that the charges inside the volume enclosed by the Gaussian box reside on an area of the plane. Hence,
(2.3.7)
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 from a uniformly charged plane in the plane:
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 is located. Note that above the plane, , while below the plane, .
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.
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.
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)
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 and the surface charge densities and . This means that the net field inside the conductor is different from the field outside the conductor.
(Figure 2.4.2)
The redistribution of charges is such that the sum of the three contributions at any point inside the conductor is
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, and hence
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)
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 everywhere inside a conductor,
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)
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 , a device that can detect a very slight amount of charge flowing from one shell to the other. When switch is thrown to the left, charge is placed on the outer shell by the battery . 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 , would be less than . More recent measurements by Williams, Faller, and Hill (1971) place at less than , a number so small that the validity of Coulomb’s law seems indisputable.
(Figure 2.4.5)
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 and the magnitude of the electric field are related by
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 crosssectional area of the cylinder are and , 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 is essentially constant over the surface enclosed, so the total charge inside the Gaussian cylinder is . Now 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 , since the cylinder is assumed to be small enough that is approximately constant over that area. From Gauss’ law,
Thus,
(Figure 2.4.6)
The infinite conducting plate in Figure 2.4.7 has a uniform surface charge density . Use Gauss’ law to find the electric field outside the plate. Compare this result with that previously calculated directly.
(Figure 2.4.7)
For this case, we use a cylindrical Gaussian surface, a side view of which is shown.
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 , so the total flux through the Gaussian surface is rather than . Then from Gauss’ law,
and the electric field outside the plate is
This result is in agreement with the result from the previous section, and consistent with the rule stated above.
Two large conducting plates carry equal and opposite charges, with a surface charge density of magnitude , as shown in Figure 2.4.8. The separation between the plates is . What is the electric field between the plates?
(Figure 2.4.8)
Note that the electric field at the surface of one plate only depends on the charge on that plate. Thus, apply with the given values.
The electric field is directed from the positive to the negative plate, as shown in the figure, and its magnitude is given by
This formula is applicable to more than just a plate. Furthermore, twoplate systems will be important later.
The isolated conducting sphere (Figure 2.4.9) has a radius and an excess charge . What is the electric field both inside and outside the sphere?
(Figure 2.4.9)
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 . To calculate , we apply Gauss’s law over a closed spherical surface of radius that is concentric with the conducting sphere.
Since is constant and on the sphere,
For , is within the conductor, so , and Gauss’s law gives
as expected inside a conductor. If , encloses the conductor so . From Gauss’s law,
The electric field of the sphere may therefore be written as
Notice that in the region , the electric field due to a charge placed on an isolated conducting sphere of radius is identical to the electric field of a point charge 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 , and we can replace the isolated charged spherical conductor by a point charge at its centre with impunity.
(Figure 2.4.10)
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 inside the cavity, then the charge separation takes place in the conductor, with amount of charge on the inside surface and a amount of charge at the outside surface (Figure 2.4.11(a)). For the same conductor with a charge 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)
If a conductor has two cavities, one of them having a charge inside it and the other a charge , the polarization of the conductor results in on the inside surface of the cavity , on the inside surface of the cavity , and 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)
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  
Electric flux through an open surface  
Electric flux through a closed surface  
Gauss’s law  
Gauss’s Law for systems with symmetry  
The magnitude of the electric field just outside the surface of a conductor 
where the notation used here is for a closed surface .
where qencqenc is the total charge inside the Gaussian surface .
2.1. Place it so that its unit normal is perpendicular to .
2.2. .
2.3 a. ; b. ; c. ; d. .
2.4. In this case, there is only . So, yes.
2.5. ; 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.
1. Discuss how would orient a planar surface of area in a uniform electric field of magnitude 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?
5. Two concentric spherical surfaces enclose a point charge . 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 that has a charge at its centre to the flux through a spherical surface of radius a with a charge 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 .
9. Discuss the similarities and differences between the gravitational field of a point mass m and the electric field of a point charge .
10. Discuss whether Gauss’s law can be applied to other forces, and if so, which ones.
11. Is the term 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.
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 crosssection have to be square? Must the end faces be on opposite sides of the sheet?
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 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 ?
19. The conductor in the preceding figure has an excess charge of . If a 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?
20. A uniform electric field of magnitude is perpendicular to a square sheet with sides 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 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 between two parallel plates where there is a constant electric field of 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 angle with the direction of the electric field. Note that this angle can also be given as .
23. The electric flux through a squareshaped area of side near a large charged sheet is found to be when the area is parallel to the plate. Find the charge density on the sheet.
24. Two large rectangular aluminum plates of area face each other with a separation of between them. The plates are charged with equal amount of opposite charges, . The charges on the plates face each other. Find the flux through a circle of radius between the plates when the normal to the circle makes an angle of with a line perpendicular to the plates. Note that this angle can also be given as .
25. A square surface of area is in a space of uniform electric field of magnitude . 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) , (b) and (c) . Note that these angles can also be given as .
26. A vector field is pointed along the axis, . (a) Find the flux of the vector field through a rectangle in the plane between and . (b) Do the same through a rectangle in the plane between and . (Leave your answer as an integral.)
27. Consider the uniform electric field . What is its electric flux through a circular area of radius that lies in the plane?
28. Repeat the previous problem, given that the circular area is (a) in the plane and (b) above the –plane.
29. An infinite charged wire with charge per unit length lies along the central axis of a cylindrical surface of radius and length . What is the flux through the surface due to the electric field of the charged wire?
30. Determine the electric flux through each surface whose crosssection is shown below.
31. Find the electric flux through the closed surface whose crosssections are shown below.
32. A point charge is located at the centre of a cube whose sides are of length . 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 is at an unspecified location inside a cube of side . Find the net electric flux though the surfaces of the cube.
34. A net flux of passes inward through the surface of a sphere of radius . (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 is placed at one of the corners of a cube of side , as shown below. Find the magnitude of the electric flux through the shaded face due to . Assume .
36. The electric flux through a cubical box on a side is . What is the total charge enclosed by the box?
37. The electric flux through a spherical surface is . What is the net charge enclosed by the surface?
38. A cube whose sides are of length is placed in a uniform electric field of magnitude 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 is distributed uniformly throughout a cubical volume whose edges are long. (a) What is the charge density in the cube? (b) What is the electric flux through a cube with edges that is concentric with the charge distribution? (c) Do the same calculation for cubes whose edges are long and long. (d) What is the electric flux through a spherical surface of radius that is also concentric with the charge distribution?
41. Recall that in the example of a uniform charged sphere, . Rewrite the answers in terms of the total charge on the sphere.
42. Suppose that the charge density of the spherical charge distribution shown in Figure 2.3.3 is for and zero for . 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 . What is the electric field at a distance from the wire?
44. A charge of is distributed uniformly throughout a spherical volume of radius . Determine the electric field due to this charge at a distance of (a) , (b) , and (c) 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 .
46. A total charge is distributed uniformly throughout a spherical shell of inner and outer radii and , respectively. Show that the electric field due to the charge is
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 charge put on a aluminum spherical ball at the following two points in space: (a) a point from the centre of the ball (an inside point) and (b) a point from the centre of the ball (an outside point).
48. A large sheet of charge has a uniform charge density of . 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 long copper rod of radius is charged with of charge and we seek electric field at a point from the centre of the rod. (b) A long copper rod of radius is charged with of charge and we seek electric field at a point from the centre of the rod. (c) A wooden rod is glued to a plastic rod to make a long rod, which is then painted with a charged paint so that one obtains a uniform charge density. The radius of each rod is , and we seek an electric field at a point that is from the centre of the rod. (d) Same rod as (c), but we seek electric field at a point that is from the centre of the rod.
50. A long silver rod of radius has a charge of on its surface. (a) Find the electric field at a point from the centre of the rod (an outside point). (b) Find the electric field at a point from the centre of the rod (an inside point).
51. The electric field at from the centre of long copper rod of radius has a magnitude 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 situated such that the rod passes through opposite sides of the cube perpendicularly?
52. A long copper cylindrical shell of inner radius and outer radius surrounds concentrically a charged long aluminum rod of radius with a charge density of . 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) , (b) , (c) , (d) , and (e) .
53. Charge is distributed uniformly with a density throughout an infinitely long cylindrical volume of radius . Show that the field of this charge distribution is directed radially with respect to the cylinder and that
54. Charge is distributed throughout a very long cylindrical volume of radius such that the charge density increases with the distance from the central axis of the cylinder according to , where is a constant. Show that the field of this charge distribution is directed radially with respect to the cylinder and that
55. The electric field from the surface of a copper ball of radius is directed toward the ball’s centre and has magnitude . How much charge is on the surface of the ball?
56. Charge is distributed throughout a spherical shell of inner radius and outer radius with a volume density given by , where is a constant. Determine the electric field due to this charge as a function of , the distance from the centre of the shell.
57. Charge is distributed throughout a spherical volume of radius with a density , where 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 with a charge of distributed uniformly throughout its volume. (a) What is the electric force exerted on an electron when it is 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 , where and are constants. What is the electric field produced by this charge distribution?
60. An uncharged conductor with an internal cavity is shown in the following figure. Use the closed surface along with Gauss’ law to show that when a charge is placed in the cavity a total charge is induced on the inner surface of the conductor. What is the charge on the outer surface of the conductor?
61. An uncharged spherical conductor of radius has two spherical cavities and of radii and , respectively as shown below. Two point charges and are placed at the centre of the two cavities by using nonconducting supports. In addition, a point charge is placed outside at a distance 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.
62. A positive point charge is placed at the angle bisector of two uncharged plane conductors that make an angle of . See below. Draw the electric field lines.
63. A long cylinder of copper of radius is charged so that it has a uniform charge per unit length on its surface of . (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 is charged with of charge. A copper spherical shell of inner radius and outer radius surrounds it. A total charge of 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 apart.
65. A long cylinder of aluminum of radius meters is charged so that it has a uniform charge per unit length on its surface of . (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, . Show that this equation is consistent with the fact that at the surface of a spherical conductor.
67. Two parallel plates on a side are given equal and opposite charges of magnitude . The plates are apart. What is the electric field at the centre of the region between the plates?
68. Two parallel conducting plates, each of crosssectional area , are apart and uncharged. If 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 . What is the electric field outside and inside the pipe? Assume the pipe has a diameter of .
70. A point charge is placed at the centre of a spherical conducting shell of inner radius and outer radius . The electric field just above the surface of the conductor is directed radially outward and has magnitude . (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 is surrounded by a concentric cylindrical shell of inner radius . The solid cylinder and the shell carry charges and , respectively. Assuming that the length of both conductors is much greater than or , determine the electric field as a function of , the distance from the common central axis of the cylinders, for (a) ; (b) ; and (c) .
73. Repeat the preceding problem, with .
74. A circular area is concentric with the origin, has radius , and lies in the plane. Calculate for .
75. (a) Calculate the electric flux through the open hemispherical surface due to the electric field (see below). (b) If the hemisphere is rotated by around the axis, what is the flux through it?
76. Suppose that the electric field of an isolated point charge were proportional to rather than . Determine the flux that passes through the surface of a sphere of radius centred at the charge. Would Gauss’s law remain valid?
77. The electric field in a region is given by where , and . What is the net charge enclosed by the shaded volume shown below?
78. Two equal and opposite charges of magnitude are located on the xaxis at the points and , as shown below. What is the net flux due to these charges through a square surface of side that lies in the 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.)
79. A fellow student calculated the flux through the square for the system in the preceding problem and got . What went wrong?
80. A piece of aluminum foil of thickness has a charge of 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 from the centre, assuming approximate planar symmetry.
81. Two pieces of aluminum foil of thickness face each other with a separation of . One of the foils has a charge of and the other has . (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 on the surface facing the other plate, and zero in between the plates. Find the electric flux through a 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 from the parallel direction. Note, this angle can also be .
83. The infinite slab between the planes defined by and contains a uniform volume charge density (see below). What is the electric field produced by this charge distribution, both inside and outside the distribution?