3.2.1  Electric Potential for a Finite Thickness, Infinite Sheet

We now want to apply what we developed at the end of the last lecture in deciding which of our two methods to use for calculating the electric potential due to a charge distribution. Let's consider the case of the infinite, charged, non-conducting sheet.

Figure 3.3: An infinite plane with volume charge density r and finite thickness t.

If we assume the sheet has finite thickness, t, and a volume charge density, r, then we know, from having done this problem before, that the electric field has a magnitude

E = rt 2e0 (3.2.1.11)

at any point outside the sheet. The electric field is directed perpendicularly away from the sheet (for a positively charged sheet). So, if we want to know the potential at point P, relative to the right surface of the sheet, then we can evaluate the potential via a straight line path from the right face of the sheet out to point P.

VP = - ó õ <® ® E   ·d ® s   = - ó õ E ds VP = -Ed (3.2.1.12)

If we considered trying to break the infinite sheet into infinitesimal charges and integrating, we would get the same answer, but the effort we would need to expend is significantly greater. Clearly this problem is best done using our previous knowledge of the properties of the electric field.

3.2.2  Potential of a Charged Ring

Consider a thin ring of plastic with a radius R and a uniformly distributed net charge -Q as shown in the figure below.

Figure 3.4: A thin ring of radius R and uniformly distributed charge -Q.

Find the following:

1. the electric field at point P at the center
2. the electric potential, relative to infinity, at point P

Solution:
We can quickly surmise the answer to the first part of the problem. By symmetry, the electric field at point P must be zero. Since we can break the circle into infinitesimal pieces, dQ, with arc-length ds, we can quickly see that every piece has another piece, diametrically opposite it, which delivers a contribution to the net field which has equal magnitude and opposite direction.

We would be unwise to be so hasty with the second part of this question however. The electric potential, after all, is evaluated by integrating over a path from infinity to point P. Even though the electric field at point P is zero, it is definitely not zero for any arbitrary position which is not at point P. Hence we have no reason to expect that we would get zero when integrating E · ds, and, in fact, we don't. In this case calculating using the E field is tedious compared to just integrating over the contributions due to each infinitesimal piece of the ring, dQ. To derive dQ, use the linear charge density method, l = Q/(2pR), where 2pR is the circumference, or total length, of the circle. So, dQ = l ds.

VP = ó õ dV = ó õ k dq R = ó õ 2pR 0  - kQ 2pR2  ds = - kQ 2pR2 (2pR) VP = - kQ R (3.2.2.13)

The answer is distinctly not zero. The lesson is, be careful! Knowing E at a point says nothing about the value of V at that point. Although we can certainly say that, in regions in which E = 0, the potential does not change.

3.2.3  Electric Potential Inside a Spherical Thin Shell Conductor

Consider a hollow metal sphere of radius R which has a net charge +Q. Find the electric potential, relative to infinity, at the center of the sphere.

Figure 3.5: A thin spherical shell of radius R and charge +Q.

Solution:
We know that a conductor must have all of its net charge sitting on a surface. In this case, its the outer surface of the sphere. Furthermore, we know the electric field is the same as for a point charge sitting at the center of the sphere for points which are outside the sphere. Inside the sphere, the electric field is zero since any Gaussian surface we draw which is completely contained inside the sphere would contain no net charge. If we set the origin of our coordinate system as being at the sphere center, then we need to integrate the electric field over a path (it might as well be straight-line since that's the easiest to integrate) from infinity to the sphere and from the sphere surface to 0 (since the electric field is different inside than outside). So, remembering that the path is opposite to the direction of the E field (radially inward for the path, radially outward for E).

VP = - ó õ ® E   ·d ® s   = - ó õ R ¥  ® E   outside  ·d ® s   - ó õ 0 R  ® E   inside  ·d ® s   = - ó õ R ¥  kQ r2  dr - 0 = kQ r ê ê ê R ¥  VP = kQ R (3.2.3.14)

3.2.4  Electric Potential Inside a Hollow Metal Sphere

Suppose we allowed the shell from the previous problem to have a finite thickness? How does that change our result for the electric potential inside the shell? Assume that we have a spherical shell with outer radius b and inner radius a as shown in figure . What is the electric potential, relative to zero at infinity, for a point P at radius r < a?

Figure 3.6: A spherical, conducting shell with inner radius a  and outer radius b. We wish to find the electric potential for a point located at r < a.

Solution:
The approach to the solution is identical to our previous example for r > b. That is to say, the potential of the outer radius of the conducting shell is

Vb = - ó õ ® E   ·d ® s   = - ó õ b ¥  kQ r2  dr = kQ b

(3.2.4.15)

How about the region a < r < b? The electric field in the bulk of a conductor must be zero, so the electric potential cannot change for this region. Hence,

Va < r < b = Vb = kQ b

(3.2.4.16)

Finally, for the region r < a we also have no electric field, hence,

Vr < a = Va < r < b = Vb = kQ b

(3.2.4.17)

Therefore, for any point inside the shell (or within the conducting material of the shell), the potential is the same as the outside surface. This includes the potential at the center of the shell, hence the result is unchanged!

3.2.5  Equipotential

We take from the previous example a potent lesson: any path for which conducting material provides a continuous link must be at the same electric potential. We can make use of this to work out problems that might seem intractable at first. For example, consider two conducting spheres with differing radii R_a and R_b  sitting on insulating stands far apart. The sphere with radius R_a has an electric charge +Q. If we connect a thin, conducting line between the spheres, then disconnect it, what are the charges on the two spheres?

Figure 3.7: Two shells on insulated stands are briefly connected by a conducting line, then disconnected.

Solution:
The conducting line brings the two spheres to the same electric potential. Since the sphere with radius R_b  was initially uncharged, bringing the two spheres to the same electric potential requires some of the charge on the R_a sphere to go to the R_b sphere. This reduces the potential on R_a and increases it on R_b until the two come to the same electric potential relative to infinity. We can calculate what this charge is by noting that the final charges must satisfy charge conservation such that

Qa + Qb = Q Þ Qa = Q - Qb

(3.2.5.18)

The final condition of the potentials being equal is easily represented provided we have the spheres separated by a considerable distance. This ensures that the electric field of one sphere does not have much influence on the charge distribution of the other sphere. For the potentials to be equal, we must have

kQa Ra = kQb Rb Þ Qa = Ra Rb Qb Q - Qb = Ra Rb Qb Þ Qb = Rb Ra + Rb Q ÞQa = Ra Ra + Rb (3.2.5.19)

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