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I have a lot of records in EM, and I know all about charge induction and Gauss's theorem for systems of conductors, nonetheless I still have a problem that I cannot face without feeling uncomfortable.

Suppose to have a hollow conducting sphere, with a pointlike charge $q$ inside, placed at a point not at the centre of the sphere. That induces an asymmetrical (but axisymmetrical) charge distribution on the inner surface of the hollow sphere; but also, a perfectly homogeneous charge distribution on the external surface. Why is this?

This is something I can understand could happen, but I miss some actual proof that it must happen. It must reside in something related to the particular symmetry of the sphere, but for me it is not enough to say that "this happens due to the spherical symmetry". Is there something that clearly forces things to happen like this?

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The metal of the conductor 'shields' the outer surface charges from the inner ones, because no macroscopic electrostatic field can exist inside the metal of the conductor.As such, the outer charges have no information about the presence of inner charges. So, the charges must exist in the form so as to make the conductor surface an equipotential(since this is the lowest energy configuration). For a sphere, this is simply uniform due to homegeneity of space. It may not be uniform for another random shape, but it always MUST be an equipotential.

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  • $\begingroup$ It is indeed the poit I was referring to. The field inside the conductor, that must be zero, is not the field for each distribution, but the total resulting from the sum. If the outer charge would be deposited there after a null field condition is reached, then I agree that the charges would be uniformly distributed. BUT these distribution settle all at the same time; and there is a precise information of the presence of inner charged that is passed to the outer ones: the total amount of charge. That is the proof that outer charges are affected by inner charges. $\endgroup$ – Matteo Lorenzini Oct 11 '16 at 10:58
  • $\begingroup$ no, the very existence of outer charges is defined by the inner ones. Further change inside, keeping the magnitude constant renders no information to the outside. $\endgroup$ – Lelouch Oct 11 '16 at 15:04
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Since there must be no field inside the metal of the sphere, the charges on the inner side arrange themselves to exactly cancel the field of the point charge q.

The charges on the outside thus don't feel any field except themselves. They arrange uniformly around the sphere.

In the end, the reason why the field is independent on the exact position of the innermost charge is that it is shielded by the metal cage.

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  • $\begingroup$ The point for me is to some extent subtler. The condition that the field inside the conductor is zero can be obtained as you stated, since in particular the inner field of a spherical shell is zero. But, in principle, coming from the contributions of all the internal charges and external charges added up, one could expect that also a different arrangement, with external charges placed in a non uniform way, could do the same job. Is there a way to rule out the latter possibility? $\endgroup$ – Matteo Lorenzini Oct 11 '16 at 10:51
  • $\begingroup$ There cannot be another configuration that kills the field inside the conductor and respects conservation of total charge. You can see this by constructing the solution to the problem in three steps. First, the inner layer of the shell charges so to neutralize the effect of the point charge, enforcing no field in the metal. Charges further out than the inner layer cannot have a part in this (you can see this with Gauss' law). Finally, the same charge that has been displaced to the inside layer (but with opposite sign) appears on the outside. All of these steps are unique. $\endgroup$ – polwel Oct 11 '16 at 11:47
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    $\begingroup$ Thank you @polwel, just to understand better, the external charges cannot have part in it only if they are uniformly distributed over the sphere, that is exactly what I want to find. Otherwise, Gauss' law only give me information about an inegral value, while locally the zero field could be granted also by external charges. Where am I wrong? $\endgroup$ – Matteo Lorenzini Oct 11 '16 at 12:10
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From what I figure you're describing a shell of metal sphere enclosing some charge. It doesn't matter if you place the charge at the centre or not. The external surface charge distribution will always be uniform. If you place charge Q inside, it'll induce -Q on the inner surface and the inner surface charge distribution will be non-uniform depending on the location of the free charge. Now if you evaluate $\oint \vec E . d\vec l = 0$ on a curve passing through the shell's 'meat' and inside of of the sphere you must get that the shell itself has no field inside. Only charge has been lost from the shell to the inner surface. The total amount of charge lost is then pulled away from the external surface and if the external surface would have non-uniform charge distribution then it means there would be a tangential current flowing from the patch having higher charge than it's surroundings until the surroundings and the patch are at equipotential and thus you can't calculate where the charge inside is located.

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This is really a symmetry argument but I think answers your question?

The spherical conducting shell is an equipotential.
A charge outside that shell can only feel the effect of the charges on the outside surface of the shell there being no electric field inside the conducting shell.

If a charge, which starts at infinity, is moved to the surface of the shell the work done must be independent of the path taken.

If the surface charge density was not uniform over the whole sphere the work done in moving the charge to the sphere would not be independent of path. More work would be done moving the charge towards a region where the surface charge density was larger.

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  • $\begingroup$ I would emphasize the point that a conductor is always equipotential, no matter which would be the charge density; for instance, if you have a pointlike charge close to a conducting plane, the plane itself is equipotential even if the charge density is not uniform. The same happens to a sphere. What I'm asking for is a very general argument, valid per se and not after complicate calculations. Maybe that the configuration with an uniform charge density is a minimum of the electrostatic energy among all the possible charge configurations? $\endgroup$ – Matteo Lorenzini Oct 12 '16 at 8:43
  • $\begingroup$ A sphere has no points and therefore ant accumulation of charge in one region would mean that the work done in bringing a positive charge to the sphere would not be independent of the path taken. $\endgroup$ – Farcher Oct 12 '16 at 9:14

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