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Consider a hollow conducting spherical shell S1 inside an irregularly shaped conducting wall S2 (in the figure). The sphere S1 is somehow given a charge +Q.

enter image description here

Will the charge distribution on S1 be uniform or not?

Here's what I've deduced till now:

  1. On the inner surface of S2, charge -Q will be induced (of course, non uniform).

  2. On the outer surface of S2, for its electrical neutrality, charge +Q appears.

This is all I've come up with so far, and what's confusing me is whether or not the induced charges on the inner surface of S2 will play a role in determining the charge distribution on S1.

Please provide a detailed explanation, and help me understand what's going on. I'm quite familiar with Electrodynamics (and physics as a whole), so place no restriction on the tools being used to explain. Not to mention, more of physics and less of math is always interesting to read!

Thanks a lot.

P.S. Points A, B and C in the diagram have no relation with my question - do not get confused.

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Your points 1. and 2. are correct. But the surface charge distributions on S1 and S2 will be non-uniform. Only if S2 was a concentric conducting sphere then the surface charge distributions both on S1 and S2 would be uniform. The reason is that for a uniform surface charge you need a constant normal surface electric field both on the surface of S1 and on the inner surface of S2. Such a constant normal surface electric field on closed conducting shells can only exist when you have a spherical symmetric solution of the Laplace equation for the potential and thus the electric field between the conductors. Due to the irregular shape of the outer conducting shell this is not possible.

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    $\begingroup$ I'm sorry for the trouble, but I still didn't really get why S2 being irregular in shape would affect S1's charge distribution, could you elaborate a bit on the connection with Laplace Equation and Boundary Conditions, if possible? $\endgroup$
    – stoic-santiago
    Commented May 10, 2018 at 16:35
  • $\begingroup$ @schrodinger_16 - In principle, it can be derived from pure symmetry considerations. But you can also argue physically in the following way: The potential on both shells is constant. Thus, for a given shape of the outer shell, there must be a constant potential difference between both shells. From this follows that in the region where the distance between the shells is smallest there must be a higher electric field, both on the surface of S1 and the inner surface of S2, than there where the distance is larger and thus the electric field is lower. Thus the S1 surface field cannot be constant. $\endgroup$
    – freecharly
    Commented May 10, 2018 at 17:07
  • $\begingroup$ @schrodinger_16 - Mathematically, you could prove it by assuming a constant potential and constant charge density (surface field) at S1. From this, disregarding the outer boundary condition on S2, follows a spherically symmetric potential and field solution of Laplace's equation where the potential is constant at concentric spheres. Thus a constant potential boundary condition at S2 can only be fulfilled if S2 is a concentric sphere. $\endgroup$
    – freecharly
    Commented May 10, 2018 at 17:32

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