Induced charge on sphere

Question

I have a conducting sphere ($radius = a$) at potential $V_0$. It is enclosed by another thin shell ($radius = b, b > a$) which has a charge density $\sigma (\theta) = \sigma_0 \cos(\theta)$ for the polar angle. I'm supposed to find the potential everywhere, noting the field discontinuities at $r = a$ and $r = b$ (this probably involves Poisson's equation), as well as the surface charge density on the inner conducting sphere.

So inside the conductor (r less than a), the potential should be constant. Additionally, integrating $\sigma (\theta)$ over all $\theta$ gives me $0$ total charge (very positive at the top of the sphere and negative at the bottom of the sphere), so the potential for $r > b$ should be constant as well because there is zero total charge inside (using Gauss's law with $Q_{enclosed} = 0$).

I'm really stuck how to do the potential in the in-between shell, as well as finding the surface charge densities (though once I know the former, the latter mat be doable via the discontinuity of the field $E_{in} - E_{out} = \sigma / 2$). Any ideas?

Answer 1

First of all, potential for $r > b$ will not be constant.

You could solve the above problem by superposition.
* Potential due to inner conducting sphere only is $V_0$ for $r < a$ and $V_0*a/r$ for $r > a$
* Potential due to outer dielectric is $\sigma_0b^3\cos(\theta)/3\epsilon_0r^2$ for $r > b$ and $\sigma_0r\cos(\theta)/3\epsilon_0$ for $r<b$ (This can be calculated by treating the charge distribution as superposition of positively and negatively charged spheres. Ref. The Feynman Lectures on Physics Chapter 6)

Potential at $r=a$ is $V_0 + \sigma_0a\cos(\theta)/3\epsilon_0$

Now, as potential of conducting sphere must be constant, charges redistribute to give a potential of $-\sigma_0a\cos(\theta)/3\epsilon_0$ at $r=a$. We have seen that $\sigma = \sigma_0\cos(\theta)$ gives potential distribution of that kind.

Thus charge distribution on conducting sphere may be taken as $4\pi\epsilon_0aV_0 - \sigma_0\cos(\theta)$

Thus Potentials come out to be

$V_0$ for $r<a$
$V_0a/r+(r^3-a^3)\sigma/3r^2\epsilon_0$ for $b>r>a$
$V_0a/r+(b^3-a^3)\sigma/3r^2\epsilon_0$ for $r>b$