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If considering a hollow conducting sphere with a surrounding uniform charge distribution, for example, it will have a constant and uniform potential throughout the inside of the hollow sphere because $\phi \sim 1/r$. But if instead there were dipoles, quadrupoles, octupoles, etc. uniformly surrounding a hollow sphere with $\phi \sim 1/r^n$ and $n$ an arbitrary integer, is the potential inside the sphere necessarily uniform everywhere?

The basic answer appears to be yes by Gauss's Law since there is no charge inside the hollow sphere. And I've seen geometric arguments for the $n = 1$ case, but are there any general proofs for arbitrary $n$?

Buzz
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Mathews24
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1 Answers1

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If the conducting sphere is solid metal, then its entire interior must be an equipotential. There will be a layer of surface charge at its surface, which cancels out the electric fields of all the external sources. This surface charge distribution can be quite complicated, but a distribution with the right properties always exists.

If the sphere is hollow (with no free charge located inside the hollow), the same surface distribution exists on the exterior surface. Because the fields of the external charges, plus the surface charge layer give exactly zero field everywhere inside the sphere, there is still a vanishing electric field everywhere inside the hollow. And if $\vec{E}=0$ in that region, the potential $V$ must be a constant over the conducting shell and its hollow interior.

Buzz
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