I know that the electric field inside a charged conducting shell is zero. However, what if we accelerate the shell uniformly? Does the charge distribution remain uniform? If not, is it possible for the new charge distribution to allow some electric field to penetrate the conducting shell?
I guess that the shell still resists the electric fields to be penetrated inside it even if the shell is accelerated, however, I cannot prove it.
The excellent answer to your question is provided by Andrew Steane in the following paper available on the RG platform: “The fields and self-force of a constantly accelerating spherical shell”
Article in Proceedings of The Royal Society A Mathematical Physical and Engineering Sciences · July 2013 DOI: 10.1098/rspa.2013.0480b
Article The fields and self-force of a constantly accelerating spherical shell
That's an interesting question and the paper Janusz Pudykiewicz provided is interesting as well, but I think misses an important component from the original question. They assume a shell with uniformly distributed charge undergoing "rigid" motion. I.e. the charges are stuck in place in the shell. Mohammad Javanshiry asked about a conductor, where charges are not fixed, but can move to screen applied fields. I don't see any mention of that in a scan of Steane's paper. In a conductor, the uniformity of the charges is directly related to the electric field and vice-versa. I would expect the charges to move to try to screen any field that their acceleration was causing. I would also have to ask what is actually causing the acceleration, since that would have implications. Is it acceleration due to an applied field vs. due to gravity or some other external force. Remember too that in a real physical medium, there's more than the excess charge at play. You have the net zero charge of all the atoms that have to re-equalize under acceleration plus that of the excess charge. Acceleration under gravity would imply all elements accelerating equally, while under an electric field, there would be some polarization of the net zero components in addition to what the excess charge would do. Finally, if we assume an external mechanical force, there would be a charge-to-charge compressive force of the acceleration pushing the shell along. And remember, accelerating a charge causes electromagnetic radiation, so there are a lot of interesting things happening there.
At any rate, I don't have the answer for you, but just a few thoughts on setting up the problem. But I would expect that yes, the charges would move to some new equilibrium state, but since they can't move forever, that new state would have to balance the applied acceleration. What the net field is inside the shell after that is the real question.
I have not read the article yet, however, I want to thank you, Michael D. Foegelle, for your remarks and clarification.
The motion of the charged conducting shell with acceleration will lead to an asymmetric distribution of the resulting charge over the shell. In this case, it is important to have an excess charge that can move. The internal charge of the contribution will not give. His field will be compensated. As a result, a field should appear inside the shell.
There is a discrepancy in the definition of rigid motion in this thread.
According to Michael D. Foegelle:
>>They assume a shell with uniformly distributed charge undergoing "rigid" motion. I.e. the charges are stuck in place in the shell. Now consider a spherical shell of charge undergoing rigid hyperbolic motion. By rigid motion we mean motion such that, at any moment, there is an instantaneous rest frame in which all parts of the body are at rest, and the physical dimensions of the body in each such frame are constant (i.e. the same in all successive instantaneous rest frames), see for example Steane (2012)
Yes, the paper assumes an arbitrary distribution of charge. I was actually mixing the OP question with the statement of the paper. The OP asks if the charge distribution would remain uniform, which implies that it began uniform. That actually may be an invalid assumption itself, depending on your definition of uniform. On a conducting sphere (matching the example picture from the paper) I would expect a uniform distribution, but on an arbitrarily shaped surface, the excess charge may not be uniform, but instead it will move to whatever is the low energy state. Regardless, the point remains that the "rigid" motion spec doesn't appear to match the idea of a conducting surface, which would not be rigid. If the motion/acceleration induces a field, it would induce a current and the charges would move.
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