In the previous sections we considered only situations in which the electric field is uniform within a region. However, often the real world is more complex than this. We need to be able to calculate changes in electric potential and electric potential energy in regions of space in which the electric field is not uniform.

Consider a situation in which the path from the initial location to the final location of interest passes through two regions of uniform but different electric field. In region 1 the electric field is , and in region 2 the electric field is , as shown in Figure 17.16. What would be the change in electric potential energy of a particle traveling along a path from location A to location B? (Assume that there is a tiny hole in the middle plate, so a particle can pass through it.)

Figure 17.16    The path from location A to location B passes through two regions in which the field is different.

In order to calculate ΔV in this situation, we need to divide the path into two pieces. Each displacement vector must be small enough that the electric field is uniform in the region through which it passes. Essentially, we have to invent a new point, which we can call C, on the boundary between the two regions, as shown in Figure 17.17.

Figure 17.17    We divide the path into two segments by inventing a point C. Each segment of the path now passes through a region of uniform electric field.

Now we can calculate the difference in potential along the two segments of the path:

From location A to location C: From location C to location B: From location A to location B:

In general, we need to add up all components (x, y, and z), so we can write the general equation as a sum:

	Potential Difference Across Several Regions

example Potential Difference Across Two Regions Suppose that from x = 0 to x = 3 the electric field is uniform and given by , and that from x = 3 to x = 5 the electric field is uniform and given by (see Figure 17.18). What is the potential difference ΔV = VC − VA?    Figure 17.18    The electric field differs in the two regions. Solution   Split the path into two parts, A to B and B to C. In each part the electric field is constant in magnitude and direction. We have the following:

In Chapter 15 we proved that inside a conductor, such as a metal object, in equilibrium, the net electric field is zero (because otherwise mobile charges would shift until they contribute a field large enough to cancel the applied field). Therefore it must be the case that in equilibrium the net electric field is zero at all locations along any path through the conductor. This means that the potential difference is zero between two locations inside the metal (Figure 17.19).

Figure 17.19    Potential difference is zero inside a metal in equilibrium.

In a conductor in equilibrium the difference in potential between any two locations is zero, because E = 0 inside the conductor.

We'll use this fact, in combination with the approach developed in this section, to examine a situation involving the potential difference across multiple regions, one of which is inside a metal object. Suppose that a capacitor with large plates and a small gap of 3 mm initially has a potential difference of 6 volts from one plate to the other (Figure 17.20).

Figure 17.20    A capacitor with large plates and small gap.

question What are the direction and magnitude of the electric field in the gap?

The direction of the electric field in the gap is toward the left, away from the positive plate and toward the negative plate. The magnitude can be found from noting that the potential difference ΔV = Es = 6 volts, so that E = (6 volts)/(0.003 m) = 2000 volts/meter.

First we need to understand the new pattern of electric field that comes about when the metal slab is inserted. The charges on the outer plates are −Q₁ and +Q₁, both distributed approximately uniformly over a large plate area A. The metal slab of course polarizes and has charges of +Q₂ and −Q₂ on its surfaces, as indicated in Figure 17.21.

The electric field inside a capacitor is approximately E = (Q/A)/ε₀, where Q is the charge on one plate and A is the area of the plate, if the plate separation s is small compared to the size of the plates.

question Since the electric field inside the metal slab must be zero, we can conclude that Q2 is equal to Q1. Why?

The plates and the surfaces of the slab have the same area A. The outer charges −Q₁ and +Q₁ produce an electric field in the metal slab E₁ = (Q₁/A)/ε₀ to the left. The inner charges +Q₂ and −Q₂ are arranged like the charges on a capacitor, so they produce an electric field in the metal slab E₂ = (Q₂/A)/ε₀ to the right. The sum of these two contributions must be zero, because the electric field inside a metal in equilibrium must be zero. Hence Q₂ is equal to Q₁.

As a result, the left pair of charges produces a field like that of a capacitor, and the right pair of charges also produces a field like that of a capacitor. The effect is that after inserting the metal slab, the electric field is still approximately 2000 V/m in the air gaps but is now zero inside the metal slab. (The fringe fields are small if the gap is small.)

question Now that we know the electric field everywhere, what are the potential differences across each of the three regions between the plates (air gap, metal slab, air gap)?

The electric field in the air gap is essentially unchanged, so ΔV inside the metal slab must be zero, because E = 0 inside the slab. The potential difference between the plates of the capacitor is now: not 6 V as it was originally. (Note that there is no such thing as “conservation of potential”; the potential difference changed when we inserted the slab.)

In some situations, the electric field in a region varies continuously. For example, the electric field due to a single point charge is different in magnitude and direction at every location in space. To find a change in electric potential between two locations in such a region, we need to take into account the varying electric field.

One approach would be to divide the path between the two locations into a finite number of steps, and use an average value of the electric field along each step to compute an approximate potential difference along that step. We could add these approximate values to get an approximate value for the potential difference along the entire path. As we make the length of each step smaller and smaller, the sum turns into an integral:

	Potential Difference in a Region of Varying Field
	which can be evaluated as the sum of three separate integrals:

To calculate the potential difference between points a and b, always begin by considering the equation

(a)   Decide whether varies in the region of interest. Draw the pattern of electric field throughout this region, making sure that you are considering the electric field not only at points a and b, but at locations in space around those points as well. (b)   Specify the origin and a set of axes for this situation. (c)   Write as a function of position using the coordinates specified by your chosen axes. (d)   Draw a path that connects points a and b. Be sure that your path starts at the initial location and ends at the final location. Based on this path and the electric field in the region, predict the sign for the potential difference associated with moving along the path. (e)   Convert to a quantity that represents a displacement along the path you have chosen in the coordinates given by the axes. (f)   Evaluate and integrate (or sum numerically). Check the sign that you obtain after performing your integration and compare it against your prediction.

Remember that you are free to choose any path that connects the two points, since the electric potential difference between two locations in space is independent of path.

Picking a path that takes into consideration the dot product and the field pattern you've already drawn could significantly simplify the problem.

We'll apply this procedure in a familiar situation: a region near a single point charge, as shown in Figure 17.23.

Figure 17.23    The electric field varies continuously along a path near a point charge.

(a)   In our drawing of the region around the point charge (Figure 17.23) we see that varies, so we need to use an integral. (b)   We pick the location of the charge as the origin, with the x axis in the direction from location A to location B. (c)   We write the field between A and B as a function of x: (d)   We choose as our path a straight line along our x axis, from A to B. Since the direction of the path from A to B is the same as the direction of the electric field in the region, the potential difference should be negative. (e)   We divide the path up into a very large number of infinitesimal steps, each of the same magnitude and direction. Instead of , we will call each in-finitesimal displacement vector . (f)

The value of x at location A is x_a, and the value of x at location B is x_b, so these are the limits of integration: We can determine the sign of this quantity by looking at both the sign of Q and the difference . In this case Q > 0 and x_b > x_a, so is negative, and ΔV is negative, as we expected in step (d).

If the point charge were negative, the potential difference would be positive; this also makes sense, since the electric field would be opposite to the direction of the path.

example Potential Difference Due to a Proton A proton is located at the origin. Location C is 1 × 10−10 m from the proton, and location D is 2 × 10−8 m from the proton, along a line radially outward. (a)   What is the potential difference VD − VC ? (b)   How much work would be required to move an electron from location C to location D? Solution   (a)   Potential difference: (b)   Work to move an electron: The least amount of work will be required if we don't change the kinetic energy of the proton.

The most general definition of potential difference is the integral expression, which can be used in any situation.

	General Definition of Potential Difference

This general expression is correct for regions in which the field is uniform, as well as for regions in which the field varies. For example, in the situation shown in Figure 17.24, where the field is uniform, The result of the integration is the same as we got in the example involving uniform field in Section 17.3.

Figure 17.24    A region of uniform electric field.

Using to calculate the potential difference along a path that is not straight, such as that shown in Figure 17.25, along which the electric field is varying in magnitude and direction, can be mathematically challenging. In many cases the best way to do this is to do it numerically, using a computer. In this text we will not ask you to do arbitrarily complex integrations of this kind analytically.

Figure 17.25    The potential difference for each step along this path is . The potential difference is the sum of all these contributions.

To establish the basic principles we will typically consider relatively simple situations. Often the electric field is uniform (same magnitude and direction) all along a straight path, which makes the calculation very simple. Another relatively simple situation is one in which the magnitude varies but not the direction along a straight path. A third simple situation that we will encounter is one in which we move along a curved path, but the electric field is always in the same direction as this curved path, and often constant in magnitude, which occurs in electric circuits. In each of these simple cases it can be quite easy to evaluate the potential difference, as we saw in the preceding exercises.

example A Disk and a Spherical Shell A thin spherical shell made of plastic carries a uniformly distributed negative charge −Q1. A thin circular disk made of glass carries a uniformly distributed positive charge +Q2. The radius R1 of the plastic spherical shell is very small compared to the large radius R2 of the glass disk. The distance from the surface of the spherical shell to the glass disk is d, and d is much smaller than R2. Find the potential difference V2 − V1. Location 1 is at the center of the plastic sphere, and location 2 is just outside the glass disk. State what approximations or simplifying assumptions you make. Solution   Alternatively, we could add the two electric field contributions, then integrate along a path from 1 to 2. Initial location: 1 Final location: 2 Path: Straight line from 1 to 2 (see Figure 17.27). according to the Superposition Principle, so    Figure 17.26    A uniformly charged plastic shell and a uniformly charged glass disk.    Figure 17.27    Electric field contributions of the shell and the disk, shown at locations along the path. This allows us to find ΔV due to the shell and ΔV due to the disk separately, then add them. Simplifying assumption: Neglect the polarization of the plastic and glass, because there is little matter in the thin shell and thin disk, so the field of the polarized molecules is negligible compared to the contributions of −Q1 and +Q2. ΔV due to shell: Put origin at center of shell. Inside the shell: Vsurface of shell − V1 = 0 because Eshell = 0 inside the shell. To the right of the shell: Check sign: As we move toward the disk, we're moving opposite to the field, so the potential should increase. Result agrees, since +Q1/R1 is the larger term. ΔV due to disk: an approximation is that since and , We could set up an integral, placing the origin at the center of the disk. However, since is approximately uniform, we recognize that the result of integrating will be Check sign: As we move toward the disk, we're moving opposite to the field, so the potential should increase. Result agrees. ΔV due to both shell and disk: