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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 $n$?

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    $\begingroup$ Is the sphere supposed to be a conductor? $\endgroup$ – Buzz Nov 13 '18 at 1:35
  • $\begingroup$ Yes, it's a perfectly conducting sphere. $\endgroup$ – Mathews24 Nov 13 '18 at 2:29
  • $\begingroup$ Since you know $n=1$ is true, perhaps you could employ a mathematical induction? $\endgroup$ – Aaron Stevens Nov 13 '18 at 3:22
  • $\begingroup$ The link in my question explains the geometric argument for $n=1$. It does not appear to be trivial to extend to different $n$, although that's essentially what I'm seeking. $\endgroup$ – Mathews24 Nov 13 '18 at 4:32
  • $\begingroup$ the Halbach array shows that with dipoles you can create non-zero internal fields; here are some pictures for cylindrical arrangement en.wikipedia.org/wiki/Halbach_array#/media/… , there is some verbiage in the same article about spherical arrangement, as well. $\endgroup$ – hyportnex Nov 13 '18 at 12:49
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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.

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  • $\begingroup$ I certainly agree based on Gauss's law, which you've explained in words. In general, an arbitrarily complicated surface charge distribution must ensure a constant potential inside. But the case I've given for $1/r^n$ just seems like a generalized form of the monopole expansion one gets for $1/r$, and for which proofs exist. I'm searching for proofs for why this holds for arbitrary $n$. $\endgroup$ – Mathews24 Nov 13 '18 at 4:38

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