# Heat produced in movement of charge from inside of shell to outside

The original question is

Two concentric metallic shell’s of radius R and 2R, out of which the inner shell is having charge Q and outer shell is uncharged. If they are connected with a conducting wire. Then what is the heat produced in the wire?

Well my idea is as soon as the wire is connected, it becomes a metallic shell. And the usual thing happens, all the charges move to the surface.

I calculate the change in potential of the shell in the two cases.

$$\Delta U$$=kQ/R-kQ/2R. We multiply it by the charge to get the change in potential energy. The answer should be $$kQ^2/2R$$. But the answer is half of that. I am missing some part of energy conservation. Surely not all of the change in potential energy gets converted to heat. Where does half of the change go then?

The energy of a conductor at potential $$V$$ is $$(1/2)QV$$, not $$QV$$.
If you imagine that you assemble the charges from infinity, the potential will grow gradually which explain the $$1/2$$.
• A capacitor is formed of two conductors in total influence. Charges on the two faces are opposite. Before the discharge, we can actually consider that we have a capacitor formed by the inner sphere of charge $Q$ and the outer sphere (radius $2R$) with charge $-Q$ on its inner face. But since the outer sphere is neutral, it also carries the charge $Q$ on its outer face. Finally, as its thickness is zero it is without effect and everything happens as if we had the inner sphere alone. After the discharge, there is a single spherical conductor with radius $2R$ and charge $Q$. – Vincent Fraticelli Jan 25 at 15:53