Discoveries

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

(Figure 1.1.1)  

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

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

(Figure 1.1.2)  

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

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

(Figure 1.1.3)  

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

(Figure 1.1.4)  

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

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

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

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

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

(Figure 1.1.5)  

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

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

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

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

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

Properties of Electric Charge

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

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

The Source of Charges: The Structure of the Atom

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

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

Since the atom as a whole is electrically neutral, the next question was to determine how the positive and negative charges are distributed within the atom. Thomson himself imagined that his electrons were embedded within a sort of positively charged paste, smeared out throughout the volume of the atom. However, in 1908, the New Zealand physicist Ernest Rutherford showed that the positive charges of the atom existed within a tiny core—called a nucleus—that took up only a very tiny fraction of the overall volume of the atom, but held over 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)  

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

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

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

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

(Figure 1.1.7)  

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

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

A Note on Terminology

As noted previously, electric charge is a property that an object can have. This is similar to how an object can have a property that we call mass, a property that we call density, a property that we call temperature, and so on. Technically, we should always say something like, “Suppose we have a particle that carries a charge of .” 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.

Conductors and Insulators

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

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

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

Charging by Induction

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

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

(Figure 1.2.2)  

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

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

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

(Figure 1.2.3)  

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

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

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

(Figure 1.2.4)  

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

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

(Figure 1.2.5)  

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

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).

Multiple Source Charges

The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat the analysis, two charges at a time. Specifically, we ask the question: Given  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)  

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

(Note that the force vector  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.

EXAMPLE 1.3.2

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The Net Force from Two Source Charges

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)  

Figure 1.3.4. Source charges and each apply a force on .

Strategy

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.

Solution

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.

Significance

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.

Defining a Field

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)  

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

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

In this respect, the electric field  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 .

The Meaning of “Field”

Recall from your studies of gravity that the word “field” in this context has a precise meaning. A field, in physics, is a physical quantity whose value depends on (is a function of) position, relative to the source of the field. In the case of the electric field, Equation 1.4.2 shows that the value of  (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 charge-bearing 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.

Superposition

Yet another experimental fact about the field is that it obeys the superposition principle. In this context, that means that we can (in principle) calculate the total electric field of many source charges by calculating the electric field of only  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.

The Direction of the Field

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

EXAMPLE 1.4.1

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The E-field of an Atom

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?

Strategy

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

(Figure 1.4.2)  

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

Solution

The electric field is calculated by

   

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.

EXAMPLE 1.4.2

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The E-Field above Two Equal 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 right-hand charge  instead of .

(Figure 1.4.3)  

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

Strategy

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

Solution

(a) By symmetry, the horizontal ()-components of  cancel (Figure 1.4.4);

   

(Figure 1.4.4)  

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

The vertical ()-component is given by

   

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

What we want to do now is replace the quantities in this expression that we don’t know (such as ), 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

(1.4.4)  

Significance

It is a very common and very useful technique in physics to check whether your answer is reasonable by evaluating it at extreme cases. In this example, we should evaluate the field expressions for the cases , , 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

(1.4.5)  

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.

Rotation of a Dipole due to an Electric Field

For now, we deal with only the simplest case: The external field is uniform in space. Suppose we have the situation depicted in Figure 1.7.1, where we denote the distance between the charges as the vector , 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:

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

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 so-called 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.

Induced Dipoles

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

(Figure 1.7.2)  

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

An important fact here is that, just as for a rotated polar molecule, the result is that the dipole moment ends up aligned parallel to the external electric field. Generally, the magnitude of an induced dipole is much smaller than that of an inherent dipole. For both kinds of dipoles, notice that once the alignment of the dipole (rotated or induced) is complete, the net effect is to decrease the total electric field  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)  

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

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

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.

Area Vector

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:

• Magnitude is equal to area ()
• Direction is along the normal to the surface (); that is, perpendicular to the surface.

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

(Figure 2.1.3)  

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

Since  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)  

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

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

Electric Flux

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

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

(Figure 2.1.5)  

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

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

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

(Figure 2.1.6)  

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

To keep track of the patches, we can number them from 1 through . 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.

EXAMPLE 2.1.1

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Flux of a Uniform Electric Field

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)  

Figure 2.1.7 Calculating the flux of through a rectangular surface.

Strategy

Apply the definition of flux: , where the definition of dot product is crucial.

Solution

1. In this case, .
2. Here, the direction of the area vector is either along the positive -axis or toward the negative -axis. Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.

Significance

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

EXAMPLE 2.1.2

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Flux of a Uniform Electric Field through a Closed Surface

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)  

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

Strategy

Apply the definition of flux: , noting that a closed surface eliminates the ambiguity in the direction of the area vector.

Solution

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 .

Significance

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

EXAMPLE 2.1.3

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Electric Flux through a Plane, Integral Method

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)  

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

Strategy

Apply , where the direction and magnitude of the electric field are constant.

Solution

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 open-surface equation, we find that the electric flux through the surface is

   

Significance

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

EXAMPLE 2.1.4

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Inhomogeneous Electric Field

What is the total flux of the electric field  through the rectangular surface shown in Figure 2.1.10?

(Figure 2.1.10)  

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

Strategy

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 x-axis, and each strip has an area .

Solution

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

   

Significance

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

Electric Field Lines Picture

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

(Figure 2.2.2)  

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

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

You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. A typical field line enters the surface at  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)  

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

Statement of Gauss’s Law

Gauss’s law generalizes this result to the case of any number of charges and any location of the charges in the space inside the closed surface. According to Gauss’s law, the flux of the electric field  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)  

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

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

If the charges are discrete point charges, then we just add them. If the charge is described by a continuous distribution, then we need to integrate appropriately to find the total charge that resides inside the enclosed volume. For example, the flux through the Gaussian surface  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)  

Figure 2.2.5 The flux through the Gaussian surface shown, due to the charge distribution, is .

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)  

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

EXAMPLE 2.2.1

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Electric Flux through Gaussian Surfaces

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

(Figure 2.2.7)  

Figure 2.2.7 Various Gaussian surfaces and charges.

Strategy

From Gauss’s law, the flux through each surface is given by , where  is the charge enclosed by that surface.

Solution

For the surfaces and charges shown, we find

1. .
2. .
3. .
4. .
5. .

Significance

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

CHECK YOUR UNDERSTANDING 2.3

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

(Figure 2.2.8)  

Figure 2.2.8 A cubical Gaussian surface with various charge distributions.

Charge Distribution with Spherical Symmetry

A charge distribution has spherical symmetry if the density of charge depends only on the distance from a point in space and not on the direction. In other words, if you rotate the system, it doesn’t look different. For instance, if a sphere of radius 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)  

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

One good way to determine whether or not your problem has spherical symmetry is to look at the charge density function in spherical coordinates, . 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.

Gaussian surface and flux calculations

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

(Figure 2.3.2)  

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

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 ,

   

Computing enclosed charge

The more interesting case is when a spherical charge distribution occupies a volume, and asking what the electric field inside the charge distribution is thus becomes relevant. In this case, the charge enclosed depends on the distance  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)  

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

If point  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.

EXAMPLE 2.3.1

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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.

Strategy

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

Solution

The charge enclosed by the Gaussian surface is given by

   

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)  

Figure 2.3.4 Electric field of a uniformly charged, non-conducting sphere increases inside the sphere to a maximum at the surface and then decreases as . Here, . The electric field is due to a spherical charge distribution of uniform charge density and total charge as a function of distance from the centre of the distribution.

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)  

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

Significance

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.

EXAMPLE 2.3.2

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Non-Uniformly Charged Sphere

A non-conducting sphere of radius  has a non-uniform 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.

Strategy

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

Solution

Since the given charge density function has only a radial dependence and no dependence on direction, we have a spherically symmetrical situation. Therefore, the magnitude of the electric field at any point is given above and the direction is radial. We just need to find the enclosed charge , 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:

   

(Figure 2.3.6)  

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

(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 YOUR UNDERSTANDING 2.4

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

Charge Distribution with Cylindrical Symmetry

A charge distribution has cylindrical symmetry if the charge density depends only upon the distance  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)  

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

Consequences of symmetry

In all cylindrically symmetrical cases, the electric field  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)  

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

Gaussian surface and flux calculation

To make use of the direction and functional dependence of the electric field, we choose a closed Gaussian surface in the shape of a cylinder with the same axis as the axis of the charge distribution. The flux through this surface of radius  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)  

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

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

   

whereas the flux through the end caps is zero because  there. Thus, the flux is

   

Computing enclosed charge

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:

   

EXAMPLE 2.3.3

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Uniformly Charged Cylindrical Shell

A very long non-conducting 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.

Strategy

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

Solution

(a) Electric field at a point outside the shell. For a point outside the cylindrical shell, the Gaussian surface is the surface of a cylinder of radius  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)  

Figure 2.3.10 A Gaussian surface surrounding a cylindrical shell.

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)  

Figure 2.3.11 A Gaussian surface within a cylindrical shell.

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

   

Significance

Notice that the result inside the shell is exactly what we should expect: No enclosed charge means zero electric field. Outside the shell, the result becomes identical to a wire with uniform charge .

CHECK YOUR UNDERSTANDING 2.5

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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.

Charge Distribution with Planar Symmetry

A planar symmetry of charge density is obtained when charges are uniformly spread over a large flat surface. In planar symmetry, all points in a plane parallel to the plane of charge are identical with respect to the charges.

Consequences of symmetry

We take the plane of the charge distribution to be the -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)  

Figure 2.3.12 The components of the electric field parallel to a plane of charges cancel out the two charges located symmetrically from the field point . Therefore, the field at any point is pointed vertically from the plane of charges. For any point and charge , we can always find a with this effect.

Gaussian surface and flux calculation

In the present case, a convenient Gaussian surface is a box, since the expected electric field points in one direction only. To keep the Gaussian box symmetrical about the plane of charges, we take it to straddle the plane of the charges, such that one face containing the field point  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)  

Figure 2.3.13 A thin charged sheet and the Gaussian box for finding the electric field at the field point . The normal to each face of the box is from inside the box to outside. On two faces of the box, the electric fields are parallel to the area vectors, and on the other four faces, the electric fields are perpendicular to the area vectors.

Let  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

(2.3.6)  

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)  

The Electric Field inside a Conductor Vanishes

If an electric field is present inside a conductor, it exerts forces on the free electrons (also called conduction electrons), which are electrons in the material that are not bound to an atom. These free electrons then accelerate. However, moving charges by definition means nonstatic conditions, contrary to our assumption. Therefore, when electrostatic equilibrium is reached, the charge is distributed in such a way that the electric field inside the conductor vanishes.

If you place a piece of a metal near a positive charge, the free electrons in the metal are attracted to the external positive charge and migrate freely toward that region. The region the electrons move to then has an excess of electrons over the protons in the atoms and the region from where the electrons have migrated has more protons than electrons. Consequently, the metal develops a negative region near the charge and a positive region at the far end (Figure 2.4.1). As we saw in the preceding chapter, this separation of equal magnitude and opposite type of electric charge is called polarization. If you remove the external charge, the electrons migrate back and neutralize the positive region.

(Figure 2.4.1)  

Figure 2.4.1 Polarization of a metallic sphere by an external point charge . The near side of the metal has an opposite surface charge compared to the far side of the metal. The sphere is said to be polarized. When you remove the external charge, the polarization of the metal also disappears.

The polarization of the metal happens only in the presence of external charges. You can think of this in terms of electric fields. The external charge creates an external electric field. When the metal is placed in the region of this electric field, the electrons and protons of the metal experience electric forces due to this external electric field, but only the conduction electrons are free to move in the metal over macroscopic distances. The movement of the conduction electrons leads to the polarization, which creates an induced electric field in addition to the external electric field (Figure 2.4.2). The net electric field is a vector sum of the fields of  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)  

Figure 2.4.2 In the presence of an external charge , the charges in a metal redistribute. The electric field at any point has three contributions, from and the induced charges and . Note that the surface charge distribution will not be uniform in this case.

The redistribution of charges is such that the sum of the three contributions at any point  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

Charge on a Conductor

An interesting property of a conductor in static equilibrium is that extra charges on the conductor end up on the outer surface of the conductor, regardless of where they originate. Figure 2.4.3 illustrates a system in which we bring an external positive charge inside the cavity of a metal and then touch it to the inside surface. Initially, the inside surface of the cavity is negatively charged and the outside surface of the conductor is positively charged. When we touch the inside surface of the cavity, the induced charge is neutralized, leaving the outside surface and the whole metal charged with a net positive charge.

(Figure 2.4.3)  

Figure 2.4.3 Electric charges on a conductor migrate to the outside surface no matter where you put them initially.

To see why this happens, note that the Gaussian surface in Figure 2.4.4 (the dashed line) follows the contour of the actual surface of the conductor and is located an infinitesimal distance within it. Since  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)  

Figure 2.4.4 The dashed line represents a Gaussian surface that is just beneath the actual surface of the conductor.

This particular property of conductors is the basis for an extremely accurate method developed by Plimpton and Lawton in 1936 to verify Gauss’s law and, correspondingly, Coulomb’s law. A sketch of their apparatus is shown in Figure 2.4.5. Two spherical shells are connected to one another through an electrometer , 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)  

Figure 2.4.5 A representation of the apparatus used by Plimpton and Lawton. Any transfer of charge between the spheres is detected by the electrometer .

The Electric Field at the Surface of a Conductor

If the electric field had a component parallel to the surface of a conductor, free charges on the surface would move, a situation contrary to the assumption of electrostatic equilibrium. Therefore, the electric field is always perpendicular to the surface of a conductor.

At any point just above the surface of a conductor, the surface charge density  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 cross-sectional 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,

(Figure 2.4.6)  

Figure 2.4.6 An infinitesimally small cylindrical Gaussian surface surrounds point , which is on the surface of the conductor. The field is perpendicular to the surface of the conductor outside the conductor and vanishes within it.

EXAMPLE 2.4.1

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Electric Field of a Conducting Plate

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)  

Figure 2.4.7 A side view of an infinite conducting plate and Gaussian cylinder with cross-sectional area .

Strategy

For this case, we use a cylindrical Gaussian surface, a side view of which is shown.

Solution

The flux calculation is similar to that for an infinite sheet of charge from the previous chapter with one major exception: The left face of the Gaussian surface is inside the conductor where , so the total flux through the Gaussian surface is rather than . Then from Gauss’ law,

   

and the electric field outside the plate is

   

Significance

This result is in agreement with the result from the previous section, and consistent with the rule stated above.

EXAMPLE 2.4.2

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Electric Field between Oppositely Charged Parallel Plates

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)  

Figure 2.4.8 The electric field between oppositely charged parallel plates. A test charge is released at the positive plate.

Strategy

Note that the electric field at the surface of one plate only depends on the charge on that plate. Thus, apply with the given values.

Solution

The electric field is directed from the positive to the negative plate, as shown in the figure, and its magnitude is given by

   

Significance

This formula is applicable to more than just a plate. Furthermore, two-plate systems will be important later.

EXAMPLE 2.4.3

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A Conducting Sphere

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)  

Figure 2.4.9 An isolated conducting sphere.

Strategy

The sphere is isolated, so its surface change distribution and the electric field of that distribution are spherically symmetrical. We can therefore represent the field as . To calculate , we apply Gauss’s law over a closed spherical surface  of radius  that is concentric with the conducting sphere.

Solution

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

   

Significance

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)  

Figure 2.4.10 Electric field of a positively charged metal sphere. The electric field inside is zero, and the electric field outside is same as the electric field of a point charge at the center, although the charge on the metal sphere is at the surface.

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)  

Figure 2.4.11 (a) A charge inside a cavity in a metal. The distribution of charges at the outer surface does not depend on how the charges are distributed at the inner surface, since the -field inside the body of the metal is zero. That magnitude of the charge on the outer surface does depend on the magnitude of the charge inside, however. (b) A charge outside a conductor containing an inner cavity. The cavity remains free of charge. The polarization of charges on the conductor happens at the surface.

If a conductor has two cavities, one of them having a charge  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)  

Figure 2.4.12 The charges induced by two equal and opposite charges in two separate cavities of a conductor. If the net charge on the cavity is nonzero, the external surface becomes charged to the amount of the net charge.

The Electron-Volt

The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects.

Figure 3.2.2 shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates, as it might be in an old-model television tube or oscilloscope. The electron gains kinetic energy that is later converted into another form—light in the television tube, for example. (Note that in terms of energy, “downhill” for the electron is “uphill” for a positive charge.) Since energy is related to voltage by , we can think of the joule as a coulomb-volt.

(Figure 3.2.2)  

Figure 3.2.2 A typical electron gun accelerates electrons using a potential difference between two separated metal plates. By conservation of energy, the kinetic energy has to equal the change in potential energy, so . The energy of the electron in electron-volts is numerically the same as the voltage between the plates. For example, a potential difference produces electrons. The conceptual construct, namely two parallel plates with a hole in one, is shown in (a), while a real electron gun is shown in (b).

An electron accelerated through a potential difference of is given an energy of . It follows that an electron accelerated through gains . A potential difference of () gives an electron an energy of (), and so on. Similarly, an ion with a double positive charge accelerated through gains of energy. These simple relationships between accelerating voltage and particle charges make the electron-volt a simple and convenient energy unit in such circumstances.

The electron-volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron-volts. For example, about of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of , it acquires an energy of () and can break up as many as of these molecules ( per molecule molecules). Nuclear decay energies are on the order of () per event and can thus produce significant biological damage.

Conservation of Energy

The total energy of a system is conserved if there is no net addition (or subtraction) due to work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, . A loss of  for a charged particle becomes an increase in its . Conservation of energy is stated in equation form as

or

where and stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

Voltage and Electric Field

So far, we have explored the relationship between voltage and energy. Now we want to explore the relationship between voltage and electric field. We will start with the general case for a non-uniform  field. Recall that our general formula for the potential energy of a test charge  at point  relative to reference point  is

When we substitute in the definition of electric field (), this becomes

Applying our definition of potential () to this potential energy, we find that, in general,

From our previous discussion of the potential energy of a charge in an electric field, the result is independent of the path chosen, and hence we can pick the integral path that is most convenient.

Consider the special case of a positive point charge  at the origin. To calculate the potential caused by  at a distance  from the origin relative to a reference of at infinity (recall that we did the same for potential energy), let and , with and use . When we evaluate the integral

for this system, we have

which simplifies to

This result,

is the standard form of the potential of a point charge. This will be explored further in the next section.

To examine another interesting special case, suppose a uniform electric field  is produced by placing a potential difference (or voltage)  across two parallel metal plates, labeled  and  (Figure 3.2.3). Examining this situation will tell us what voltage is needed to produce a certain electric field strength. It will also reveal a more fundamental relationship between electric potential and electric field.

(Figure 3.2.3)  

Figure 3.2.3 The relationship between and for parallel conducting plates is . (Note that in magnitude. For a charge that is moved from plate at higher potential to plate at lower potential, a minus sign needs to be included as follows: .)

From a physicist’s point of view, either  or  can be used to describe any interaction between charges. However,  is a scalar quantity and has no direction, whereas  is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field, a scalar quantity, is represented by .) The relationship between  and  is revealed by calculating the work done by the electric force in moving a charge from point  to point . But, as noted earlier, arbitrary charge distributions require calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in Figure 3.2.3 to move a positive charge  from , the positive plate, higher potential, to , the negative plate, lower potential, is

The potential difference between points  and  is

Entering this into the expression for work yields

Work is ; here , since the path is parallel to the field. Thus, . Since , we see that .

Substituting this expression for work into the previous equation gives

The charge cancels, so we obtain for the voltage between points  and

where  is the distance from  to , or the distance between the plates in Figure 3.2.3. Note that this equation implies that the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus, the following relation among units is valid:

Furthermore, we may extend this to the integral form. Substituting Equation 3.2.2 into our definition for the potential difference between points  and , we obtain

which simplifies to

As a demonstration, from this we may calculate the potential difference between two points ( and ) equidistant from a point charge  at the origin, as shown in Figure 3.2.4.

(Figure 3.2.4)  

Figure 3.2.4 The arc for calculating the potential difference between two points that are equidistant from a point charge at the origin.

To do this, we integrate around an arc of the circle of constant radius between  and , which means we let , while using . Thus,

for this system becomes

However,  and therefore

This result, that there is no difference in potential along a constant radius from a point charge, will come in handy when we map potentials.

EXAMPLE 3.2.4

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What Is the Highest Voltage Possible between Two Plates?

Dry air can support a maximum electric field strength of about . Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by of dry air?

Strategy

We are given the maximum electric field  between the plates and the distance  between them. We can use the equation  to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

   

Entering the given values for  and  gives

   

or

   

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

Significance

One of the implications of this result is that it takes about to make a spark jump across a () gap, or for a spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage can cause a spark if there are spines on the surface, since sharp points have larger field strengths than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up with static electricity on dry days (Figure 3.2.5).

(Figure 3.2.5)  

Figure 3.2.5 A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). This form of detector is now archaic and no longer in use except for demonstration purposes. (credit b: modification of work by Jack Collins)

EXAMPLE 3.2.6

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Calculating Potential of a Point Charge

Given a point charge  at the origin, calculate the potential difference between point  a distance from , and  a distance  from , where the two points have an angle of  between them (Figure 3.2.6).

(Figure 3.2.6)  

Figure 3.2.6 Find the difference in potential between and .

Strategy

Do this in two steps. The first step is to use  and let  and , with  and . Then perform the integral. The second step is to integrate  around an arc of constant radius , which means we let  with limits , still using . Then add the two results together.

Solution

For the first part,  for this system becomes  which computes to

   

For the second step,  becomes, but  and therefore . Adding the two parts together, we get .

Significance

We have demonstrated the use of the integral form of the potential difference to obtain a numerical result. Notice that, in this particular system, we could have also used the formula for the potential due to a point charge at the two points and simply taken the difference.

Before presenting problems involving electrostatics, we suggest a problem-solving strategy to follow for this topic.

Systems of Multiple Point Charges

Just as the electric field obeys a superposition principle, so does the electric potential. Consider a system consisting of  charges . What is the net electric potential  at a space point  from these charges? Each of these charges is a source charge that produces its own electric potential at point , independent of whatever other changes may be doing. Let  be the electric potentials at  produced by the charges  respectively. Then, the net electric potential  at that point is equal to the sum of these individual electric potentials. You can easily show this by calculating the potential energy of a test charge when you bring the test charge from the reference point at infinity to point :

Note that electric potential follows the same principle of superposition as electric field and electric potential energy. To show this more explicitly, note that a test charge  at the point  in space has distances of  from the  charges fixed in space above, as shown in Figure 3.3.2. Using our formula for the potential of a point charge for each of these (assumed to be point) charges, we find that

Therefore, the electric potential energy of the test charge is

which is the same as the work to bring the test charge into the system, as found in the first section of the chapter.

(Figure 3.3.2)  

Figure 3.3.2 Notation for direct distances from charges to a space point .