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Recently I was taught that in a conductor charge resides only on the surface of the conductor and the net charge inside a conductor is zero and so is the net electric field.

But what if the conductor had two surfaces like in a spherical conducting shell?

Suppose we provide a charge 'Q' to a spherical conducting shell. Is the charge on the inner surface of the shell equal to zero or is the charge divided equally among the two surfaces(Q/2 on the outer surface and Q/2 on the inner surface)? I'm in a dilemma.

NOTE: Initially There is no charge inside the cavity of the shell.

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2 Answers 2

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Charges repel one another and will get as far apart as they can. Since the shell is a conductor, nothing is stopping them from moving to the outer surface which is as far apart as they can get from one another.

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Unless there are charges in the interior of the conducting shell, since the electric field in a conductor must be zero at steady state, Gauss' law requires that there be no charge on the inner surface of the conductor. All of the charge will reside on the outer surface.

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