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Consider two independent conducting shells (not thin shells! i.e., their internal and external surfaces do not overlap.) whose shapes of external surfaces are identical but internal surfaces are not necessarily identical. They both enclose a point charge $q$ with the same quantity and sign but they may be stationed in different positions as to the two shells. See below.
Possible shapes of the shells and positions of the point charge. Pardon my poor drawing.

Note that in fact the two systems are completely independent (i.e., far apart), although they look "close" shown in my drawing. And of course, there is nothing else in the surroundings!
Obviously, on both shells' internal surface there will be the same quantity of induced charge $-q$, which, in turn, will induce charge $q$ on both external surfaces. Subsequently, the induced charge distributed on their external surfaces will give rise to the electric field exterior to the external surfaces.
For the two systems the only things they have in common are the shape of the external surface and the quantity and sign of the enclosed point charge. Now the question is, do they necessarily share the same electric field exterior to their external surfaces?
I know if the shells are connected to the ground the answer will be yes. But what if they are NOT connected to the ground?

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If they are conducting shells then they charge distribution on the outer shell will be absolutely independent of the position/shape of the inner surface.

There is a similar question and answer here which is about the position of a single charge inside a sphere - the reasoning is exactly the same here, though the problem is slightly different.

First point is that free electrons in the shell will move so that there is no electric field inside the material the shell is made out of.

Next point is that the charge on the inside of the shell will move depending on the shape of the inner shell.

Final point is that the charge on the exterior of the shell will move just to minimize its energy on the external surface - it will not be effected by the inside because there is no electric field that passes through the conducting material from the inner surface to the exterior.

I am not sure about connection to ground - but if the shells are isolated and they have no net charge and the total charge inside each of them is the same then the external electric fields will be the same. I guess if the shells were connected to ground then the potentials of the outside of the shells would be the same (I guess $0 V$) and any charge on the exterior would be distributed in the same way.

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  • $\begingroup$ Well, thank you but I have to say that I doubt there is a major difference between the question your provided and mine. It is definitely true that for a sphere the charge on the external surface is always evenly distributed however the enclosed point charge is placed. But a sphere shell is so special in its perfect symmetry, so I think it is not convincing, at least not from the conclusion in so special a case, to say it is also true for a general case. $\endgroup$ – Vim Jan 9 '15 at 2:22
  • $\begingroup$ @Vim what I mean is that the steps in the reasonning are the same in both questions. The key point is that there is no electric field in the conducting shell so that the external charge distribution cannot be affected by the charge distribution on the internal surface - this is true here and for the special case of the sphere. $\endgroup$ – tom Jan 9 '15 at 2:31
  • $\begingroup$ Oh I see.. I'm so stupid that I haven't realized that until just now.. orz.. <br/> $\endgroup$ – Vim Jan 9 '15 at 2:58
  • $\begingroup$ your proof is convincing. I guess what you really mean is that in one case the charge on the external surface has already achieved a distribution such that its potential is minimized, and since the point charge together with the induced charge on the internal surface ALWAYS contributes zero to the field exterior to the internal surface ,thus even after the change of the "internal distribution" there is actually no change in the force acting on the charge on the external surface and naturally it doesn't not have to take trouble changing its current distribution, do I follow you? $\endgroup$ – Vim Jan 9 '15 at 3:06
  • $\begingroup$ @Vim - yes - that is correct $\endgroup$ – tom Jan 9 '15 at 3:17

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