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 $1/2$.

You can also calculate the energy stored in the electric field.

  • $\begingroup$ So initially it was a capacitor, hence the potential energy is 1/2QV. After connecting it ceases to be a capacitor, loses all energy as heat. Hence the answer. Is it a correct interpretation of the situation? $\endgroup$ – HitchHiker224 Jan 25 at 13:51
  • $\begingroup$ I would say that it is still a conductor, but with a different radius, bigger, and therefore a greater capacity ? (Sorry for my english : conductor in equilibrium is different from condensator !) $\endgroup$ – Vincent Fraticelli Jan 25 at 14:36
  • $\begingroup$ Assuming that, can you still treat the two conductors (also I am a bit confused about your differentiation of conductor and condensator) the same. The first one then is a condenser of two concentric spherical shell for which the formula is different. The moment they are connected, the charges flow outside and it becomes one simple shell. Actually putting the values in the two formulas do give the correct answer. But were you treating them differentally? $\endgroup$ – HitchHiker224 Jan 25 at 15:35
  • $\begingroup$ 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 $Q$. $\endgroup$ – Vincent Fraticelli Jan 25 at 15:53
  • $\begingroup$ Right. You made it absolutely clear. I learnt a lot.Thank you. $\endgroup$ – HitchHiker224 Jan 25 at 16:04

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