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I'm reading Sherwood and Chabay's brilliant textbook Matter and Interactions, in particular the section that deals with how the surface charges in an electric circuit distribute themselves to generate the electric field within the wire.

One question left unanswered, however, is why the generated electric field does not affect the surface charges themselves, only the electrons flowing through the wire.

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In the above picture, wouldn't the charges on the rings also be affected by each other and the other ring?

1110101001
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2 Answers2

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It is possible that the surface charges are pinned at sites on the surface, but it is also possible that they are mobile. Even if they are mobile, their contribution to the current is infinitesimal, because I = JA = sigmaEA, and the cross-sectional area associated with the surface charges is completely negligible compared to the rest of the cross-sectional area. So whether the surface charges are mobile or pinned is irrelevant.

You might find it interesting to view this VPython program in your browser (thanks to the new GlowScript version of VPython found at glowscript.org):

http://tinyurl.com/SurfaceCharge

The surface charge distributions were calculated by a charge-field relaxation method described in the "Articles and talks" section of matterandinteractions.org. The VPython programs let you view these distributions and interactively explore the net field everywhere.

Bruce Sherwood, co-author with Ruth Chabay of the Matter & Interactions textbook.

user3786782
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The number of electrons on the surface is EXTREMELY small compared to the number of electrons in the wire. Think of it this way: In a metal there is a "sea" of electrons that are not bound to atoms but are free to roam. If this sea expands by an infinitesimal amount, there will be a small (but essential) number of electrons on the surface, and the interior of the wire will remain very nearly neutral. The electrons that move to the surface do NOT travel a long distance to get there; there were already LOTS of free electrons right next to the surface. Sorry, but I don't understand your second set of questions. I strongly recommend that you study the paper whose link is given above.

user356727
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