Self-Energy of a Conducting Shell based on Surface Charge Distribution

Question

If a charge $q$ is given to a shell of radius $r$ (conducting or non-conducting) it's self-energy is $kq^2/2r$, $k$ being Coulomb's constant.

But if the shell is conducting and the charge on the inner surface is $-q$ and on the outer surface is $2q$, then should the shell be considered a combination of two shells, one of charge $-q$ and the other of $2q$? In this case the self energy turns out to be $5kq^2/2r$.

But if we only consider the net charge on the shell, the energy is still $kq^2/2r$.

Which one is correct?

Answer 1

If the shell is conducting, it cannot have charge separation on inner and outer surfaces. This is possible, however, for a non-conducting shell. In this case, the total energy still remains $kq^2/2r$, since there is an additional (negative) contribution from the interaction between the negative and positive charges. For a thin spherical non-conducting shell with charge $-q$ on the inner surface and charge $+2q$ on the outer surface, the total energy is given by

\begin{align} E_\text{total} &= E_\text{outer} + E_\text{inner} + E_\text{interaction} \\ &= \frac{k(2q)^2}{2r} + \frac{k(-q)^2}{2r} + \frac{k(-q)(2q)}{r} \\ &= \frac{kq^2}{2r} \end{align}