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A solid conducting sphere carrying charge Q is enclosed by an initially uncharged conducting hollow spherical shell. The potential difference between the surface of the inner solid sphere and the outer surface of the shell is V. If the shell is now given an additional charge of -4Q, what is the new potential difference between those same two surfaces?

1. V
2. 2V
3. -2V
4. 4V

Correct answer: V

Solution

The potential difference between the inner sphere (radius r1) and the outer surface of the shell (radius r2) comes only from the field in the gap between them, which by Gauss's law depends only on the charge enclosed = Q on the inner sphere. V = (Q/(4*pi*epsilon0))*(1/r1 - 1/r2). Charge placed on the shell distributes on its surfaces but contributes equally to the potential of both the inner sphere and the shell's outer surface, so it does not change their difference. Hence the new potential difference is still V.

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