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Proof that all excess-charge resides at surface of perfect-conductor

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Use Gauss’s theorem and [tex]\oint {{\bf{E}} \bullet d{\bf{l}}} \equiv 0[/tex] to prove: any excess charge placed on a conductor must lie entirely on its surface. (A conductor, by definition, contains charges capable of moving freely under action of applied electric fields.


    2. Relevant equations

    Gauss's Law and [tex]\oint {{\bf{E}} \bullet d{\bf{l}}} \equiv 0[/tex]

    3. The attempt at a solution

    Consider an amorphous body of charge in some perfect-conductor, arranged arbitrarily. Do not yet assume that all charge resides at the surface. Associated with this body is a closed surface “S” defining a volume “V”:

    [draws figure of blobby volume, and blobby charge inside]

    Now: consider applying Gauss’s Law to a Gaussian sphere of radius “r” enclosing just a tip of this Q blob of charge:

    [same two blobs as before, except Gaussian sphere enclosing part of the charge...key detail is that some charge is inside and some charge is outside]

    The force on the blue-coloured bits of charge outside the Gaussian surface…let’s call the amount of charge Qout. Thus: the enclosed amount of charge is Q – Qout, and we use that in Gauss’s Law:

    [tex]\int {{\bf{\vec E}} \bullet d{\bf{\vec A}}} = \frac{{{Q_{encl}}}}{{{\varepsilon _0}}} = \frac{{Q - {Q_{out}}}}{{{\varepsilon _0}}} \to {\bf{\vec E}} = \frac{{Q - {Q_{out}}}}{{4\pi {\varepsilon _0}{r^2}}}[/tex]

    Suppose we are dealing with positive charge inside, for simplicity. Then: it is always true that:

    [tex]Q - {Q_{out}} > 0[/tex]

    Therefore: the electric field always points in this direction:

    [tex]{\bf{\vec E}} = \left| {{\bf{\vec E}}} \right|( + {\bf{\hat r}})[/tex]

    The force on any element of charge, then, is given by the usual F = qE in differential form (but not in preparation for integration...just for emphasizing the arbitrariness of the dQ that is part of Q[out] we put outside our Gauss-sphere):
    [itex]d{\bf{\vec F}} = dQ \cdot {\bf{\vec E}} = dQ \cdot \left| {{\bf{\vec E}}} \right|( + {\bf{\hat r}})[/itex]

    …which is always +r-ward…that is, outward radially. Finally: since electromagnetism is invariant under charge-inversion, we can make the same conclusion no matter the sign of “Q” and “Qout”.

    ...But perhaps that only proves that charge begins to diffuse outwards? What happens after the process continues a bit, and perhaps we get this Gaussian hollow-sphere? A skeptic of my proof could think of an infinite array of charge-geometries that are not "all at the surface", and try to thwart these conclusions. I don't know how to design a proof to handle all of the infinitude of such thwarting geometries.
     
  2. jcsd
  3. Aug 25, 2010 #2

    kuruman

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    Step 1: Use Gauss's Law to show that qencl. = 0

    Of course, this proves that the sum of all the charges inside the conductor is zero, not that there is no charge whatsoever inside. So you go to the next step.

    Step 2: Assume that you have charge +q somewhere inside and charge -q somewhere else inside. Use the the fact that the conductor is an equipotential to show that the line integral from one charge to the other cannot be zero in that case.
     
  4. Aug 25, 2010 #3
    Wait....all this stuff you're saying suggests that my conductor is charge-neutral. It has an excess charge. I mean, I could assume that I have +q somewhere, and -q somewhere else, but beyond that there is an excess +/- Q[encl] that is not equal to zero. ....Right?
     
  5. Aug 25, 2010 #4

    kuruman

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    No such suggestion. In step 1 your Gaussian surface is parallel to the surface of the conductor at infinitesimal distance ε below it. What is the integral

    [tex]\int\vec{E}\cdot d \vec{A}[/tex]

    under the circumstances?
     
  6. Aug 25, 2010 #5
    [tex]\int\vec{E}\cdot d \vec{A} = 0 [/tex]

    ...because Q[encl] = 0; the charge enclosed by a Gaussian surface just below that charged-skin of the conductor would be zero.

    Alright, so we are on the same page: a conductor with no excess charge.

    However: I'm worried that the integral above is zero only by assuming what I'm trying to prove. :-|
     
  7. Aug 25, 2010 #6

    kuruman

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    You are assuming what you are trying to prove. The integral is zero because the field inside the conductor (and therefore on the Gaussian surface) is zero. Therefore the charge enclosed by the surface is zero. In the limit ε→0, any net charge on the conductor has to be on the surface. See how it works?
     
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