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Electric Charge Problem

  1. Feb 26, 2010 #1
    Hey. I am trying to solve the following problem:

    Given a spherical shell with radius a and charge density sigma. If a very small patch of this surface were removed, leaving a small hole, what is the electric field just above and just below this hole?

    For the electric field below the hole, I think the answer is zero, by gauss' law. Since the charge enclosed would be zero (same as if the hole were not present).

    However, I'm not sure how I would go about solving for the electric field above the hole. I initially thought that I could simply calculate what the field would be if the hole was not there, and then take this result and subtract the electric field of a small disk of radius a (that would represent the hole) but I don't seem to get the right answer.

    And help would be appreciated! Thanks!
  2. jcsd
  3. Feb 26, 2010 #2


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    Hello mahorka,

    Welcome to Physics Forums!

    Have you studied electric potential yet? This problem will be a lot easier if you start by modeling the potential (the potential can be shown to be more or less unchanged whether the small hole is there or not). Then use its gradient to obtain the electric field.

    If you don't have the luxury of using potentials, the only way I can think of doing this involves some pretty nasty calculus integrating over the surface charge, treating the surface charges as infinitely small point charges. The thing that makes this tough is you don't get to use the normal equations since spherical symmetry is lost, and the point of interest is not at the origin, but instead somewhere near the surface of the shell. This way of solving the problem is possible, yet brutal.

    [Edit]: Btw, I suspect the problem might simply be asking for a verbal explanation rather than an explicit proof. Using the concept of electric potential may allow you to verbally explain what impact the hole would have, if any (without having to show the math). Or you could show the math. But potentials will make it easier either way.
    Last edited: Feb 26, 2010
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