# Homework Help: Three Conducting Shells

1. Sep 27, 2009

### psholtz

I have three concentric metal, conducting spheres, of radii a < x < b.

The spheres a and b are fixed, while x can vary over the range from a to b.

Spheres a and b are also grounded, while sphere x is maintained at a constant potential V with respect to these two spheres.

The charge on sphere x will distribute itself partly on the "inner" surface of the sphere (that facing a) and partly on the "outer" surface of the sphere (that facing b). I would like to calculate what fraction of charge distributes itself on the inside and outside surfaces of sphere x as a function of x (i.e., as x varies from a to b).

Some charge will be induced on sphere a. Denote the surface density of this charge as $$\sigma_a$$. We have:

$$Q_a = 4\pi a^2 \sigma_a$$

and in the region between a and x, the electric field is given by:

$$E = \frac{4\pi a^2\sigma_a}{r^2}$$

while the potential is given by:

$$\phi = -\frac{4\pi a^2 \sigma_a}{r}$$

From which we get that the potential difference between the two spheres is:

$$\frac{4\pi a^2 \sigma_a (x-a)}{xa} = \frac{4\pi a \sigma_a (x-a)}{x} = V$$

Since we're holding the potential difference V constant, and anticipating that the surface charge density changes as a function of time, we can express this (perhaps more clearly) as:

$$V = \frac{4\pi a (x-a)\sigma_a(x) }{x}$$

Solving for the surface charge density on the inner sphere as a function of x and V, we have:

$$\sigma_a(x) = \frac{Vx}{4\pi a (x-a)}$$

There are other components to this problem, and other reasons I have for wanting to solve this particular problem, but even here the problem that I'm running into raises its head: that is, if we look at the surface charge density when x=a, the answer diverges. In other words, $$\sigma_a(a) \rightarrow \infty$$.

This doesn't seem correct to me.. rather it seems that the surface charge density on the inner sphere should remain finite, in such a way $$4\pi a^2 \sigma_a(a) = Q$$, where Q is the total charge on sphere x.

Is there something wrong in my reasoning here?

2. Sep 28, 2009

### kuruman

You are correct being bothered by what happens when x = a, but you shouldn't be and there is nothing wrong with your reasoning. Remember where you started from. The potential on a sphere of radius x is fixed at the value V and the potential on a sphere of radius a is also fixed at the value zero. If you say x = a you are demanding that the potential be V and zero in the same region of space at the same time. That is unphysical.

Now look what happens as x approaches a. The charge density must increase because for the same potential difference you have a higher and higher electric field in the gap region. Another way to look at it is this: Suppose you have a parallel plate capacitor connected to a battery. As you bring the plates closer together, the charge on them increases because the capacitance increases. In the limit the plate separation goes to zero, the charge becomes infinite. Same thing here except that you have a spherical, not parallel plate, capacitor.