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Let's say you have a point charge inside a conducting shell with an inner radius of 5cm and an outer radius of $7cm$. The point charge has a chard of $-4C$ and the shell a charge of $6C$. This means that the $-4C$ point charge attracts $4C$ worth of charge from the shell to the shell's inner surface. This leaves $2C$ of charge on the shell's outer surface. If you use the Gaussian-surface method at a radius of $6cm$, the Gaussian surface encloses both the $-4C$ point charge and the shell's $4C$ inner surface, so the net enclosed charge is $0C$, and therefore there is no electric field inside the shell (between the inner surface and outer surface).

But is there a potential there? My first intuition was "no", but then I was thinking that if you put a positive point charge there, the positive charge would join the other positive charges on the shell's outer surface, which makes it seem there'd have to be some potential energy being converted to kinetic energy somehow...

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A conducting body can have a potential, and it need not be zero. Potential can be arbitrarily set, depending upon your reference potential. The only difference in the tratement of conducting bodies is that they must be equipotential, i.e., they must have constant potential at all points inside them (but not necessarily points inside cavities).

The potential of a metal shell due to its own charge $q_1$ is $\frac{kq_1}{r_{shell}}$ If you add a point charge $q_2$ at the center, then the potential becomes $\frac{kq_1}{r_{shell}} + \frac{kq_2}{r_{shell}}$.

Remember, potential at a point is can be defined with respect to the work required to get a test charge there from infinity. If the field at a point is zero, that doesn't imply that the field is zero all the way to infinity. It just means that you can jiggle a test charge in the neighborhood of that point without doing work.

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I know that a conducting body can have potential. I'm not sure if the one in my example does though. I understand your last paragraph, but if you can jiggle the particle at that point without doing electrical work, is the potential in that neighborhood zero? – trusktr Feb 21 '12 at 3:36
As I said, potential is relative. You can arbitrarily set potential at one point, and ise this as a reference. We usually set potential at infinity as 0, but there's nothing special about this. If you jiggle a charge, potential DIFFERENCE is zero in that area (i.e. V is constant) – Manishearth Feb 21 '12 at 4:01
And your example has a potential (relative to infinity) of $200k/7$. – Manishearth Feb 21 '12 at 4:02
Sweet. Thanks for explaining! – trusktr Feb 24 '12 at 6:48

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