(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Proof that all excess-charge resides at surface of perfect-conductor

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