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If an isolated, charged spherical shell has a uniform charge distribution, the electric field everywhere inside it is 0, by Gauss' Law. Is the converse true? That is, given an isolated, charged spherical shell such that the electric field everywhere inside it is 0, must the charge distribution on the shell be uniform? I suspect measure theory is involved, and if the statement is false, the counterexample will probably deal with non-measurable sets. Note that I am asking only about spherical shells. Thanks!

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Yes, this is guaranteed by the uniqueness theorem for Poisson's Equation, and in fact is more general than spherically charged shells.

However, as another answer indicates, that if there are other charges present elsewhere, the charge on the sphere will shift to cancel off the field interior of the conductor.

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  • $\begingroup$ So given the electric field in a region of space, there can be but one charge distribution in that region? $\endgroup$ – Davey Apr 22 '16 at 3:15
  • $\begingroup$ Speaking to a physicist, yes. To a mathematician... I don't know :) $\endgroup$ – anon01 Apr 22 '16 at 3:17
  • $\begingroup$ @Davey oh, and this is assuming electrostatics. The same does not hold for electrodynamics. $\endgroup$ – anon01 Apr 22 '16 at 3:24
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    $\begingroup$ @Davey Not quite, but almost, yes. Look up "uniqueness theorem", there are certain boundary conditions involved to guarantee a unique solution. You should actually read the proof, it is easy to follow and I think it's lovely. $\endgroup$ – Michael Angelo Apr 22 '16 at 8:15
  • $\begingroup$ I see... So is the fact that the E field is 0 along the boundary a Neumann boundary condition? $\endgroup$ – Davey Apr 22 '16 at 13:14
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When there are no external fields the charge must be distributed uniformly. If you have an external field and the sphere is made of conducting material then it will act as a Faraday cage and the charges will distribute themselves to cancel the field inside the sphere, leading to a nonuniform charge distribution on the surface with a 0 field inside the sphere.

In all cases no field in the interior of the sphere implies a constant potential throughout it, since, by definition, the field is the gradient of the potential. Therefore there can be no potential differences along the surface of sphere, and thus the tangential field component must be 0 everywhere on the surface.

In the absence of an external field this implies a uniform charge distribution on the surface. With an external field the charges need to be distributed such that the combined potential on the surface is constant so the tangential component of the field due to the charges exactly cancels that of the external field on the entire surface.

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We can assume w.l.o.g. that the electric potential

$$\left. \Phi(r,\theta,\varphi) \right|_{r<r_0}~=~0$$

vanishes in the interior. As already argued in Daniel Mahler's answer, a surface charge distribution $\rho$ with support at $r=r_0>0$ is far from unique. In fact, the reader may check that any electric potential of the form

$$ \Phi(r,\theta,\varphi) ~=~ H(r\!-\! r_0) \underbrace{\sum_{m\ell} \left(A_{ml} r^{\ell} +B_{m\ell}r^{-\ell-1} \right) Y_{m\ell}(\theta,\varphi)}_{\text{general solution to Laplace equation}} $$

leads to a surface charge distribution $\rho$ with support at $r=r_0$ via Poisson's equation. Here $H$ and $Y_{m\ell}$ denote the Heaviside step function and the spherical harmonics, respectively.

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