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Consider a closed surface $\Sigma$ which bounds a solid $\Omega$ in ${\mathbb R}^3$. Assume some electric charges, say totally $Q$, is distributed on $\Sigma$ and reaches an "equilibrium" state. In this situation the electric field inside $\Omega$ should be $0$.

What can we say about the density $\rho$ of the charge in this equilibrium distribution? Is there a simple rule that relates $\rho$ to the geometry /curvature of $\Sigma$ (e.g. something like, $\rho$ proportional to the pointwise norm of the second fundamental form - but this is just speculation)? For example, if $\Sigma$ is a round sphere, then it is well known that $\rho$ is constant.

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    $\begingroup$ There is a good bibiography here: arxiv.org/pdf/1509.09252.pdf $\endgroup$ May 10, 2017 at 6:52
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    $\begingroup$ This is called the equilibrium distribution, and it was very much studied. There is no simple expression in terms of geometry of $\Omega$. Of course it is completely determined by this geometry but the connection is complicated, and most results have the form of inequalities rather than equalities. $\endgroup$ May 10, 2017 at 7:21
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    $\begingroup$ Why on earth would some want to close this question? This is a classic question (not completely solved as far as I can see) that dates back to Maxwell and to an editor's note by J.J. Thomson---the Nobel prize winner who discovered the electron---in Maxwell's treatise. It has been studied for a long time and is a perfectly valid question. $\endgroup$ May 10, 2017 at 9:43
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    $\begingroup$ This question is related, and you might find Henry Cohn's answer there helpful: mathoverflow.net/questions/80731/… . $\endgroup$
    – j.c.
    May 10, 2017 at 11:30
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    $\begingroup$ The charge density cannot be a function of the second fundamental form, because the problem is global: changing the geometry of the surface in a neighborhood of a point has an effect on the charge distribution everywhere. $\endgroup$ May 10, 2017 at 11:53

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