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I Potential on the inner surface of a spherical shell

  1. Aug 27, 2016 #1
    Is there a potential on the inner surface of a charged spherical shell?

    I know that there is no electric field on the inner surface, as shown by Gauss's Law, but that isn't enough information to say that the potential (V) there is zero since E = dV/dr, so V could be a nonzero constant.

    If there is a potential on the inner surface, then I must clarify that what I'm curious about is what I'll call the "effective potential." That is, given this spherical geometry, for every point on the inner surface, there is a point of equal potential directly across from it. The closer these points are to each other (i.e., the smaller the sphere's radius), the smaller the effective potential at either point becomes, right? So how would one calculate this radius-dependent effective potential? Is there a formula?
     
  2. jcsd
  3. Aug 27, 2016 #2

    Dale

    Staff: Mentor

    The potential is not uniquely defined. You can do a gauge transformation and leave the fields and all measurable effects the same. So this means that you can choose a gauge so that the potential can be any value you like inside the shell.
     
  4. Aug 28, 2016 #3
    Interesting. So in other words, it wouldn't matter for any practical purposes. All measurable effects would be the same whether there were a potential there or not.
     
  5. Aug 28, 2016 #4

    Dale

    Staff: Mentor

    Yes, exactly. In fact, usually the potential is chosen to make the math simpler.
     
  6. Aug 28, 2016 #5
    Actually the phrase (or question) "is there a potential" is meaningless.
    Even assuming that the potential is zero, there is still a potential. One with value zero.
    There is no way to have a situation in which there is no potential, is there?
     
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