Gaussian surface surrounding only a proton inside a conductor

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RMZ
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Homework Statement


I understand why, using Gauss's law, the net charge within a conductor should be zero at any point. However, when I try making a Gaussian surface that is so small so as to enclose a single proton, I cannot see why the enclosed charge should be zero for that situation as well, which seems to throw off everything.

Homework Equations


Electric Flux through a Gaussian surface = Qenc/epsilonnought

The Attempt at a Solution


The electric field at any point on the Gaussian surface should be zero, which means that phiE
=0, which should mean that the enclosed total charge = 0. But if only a proton is within this Gaussian surface how is this possible?
 
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So should I think about the charge in a conductor in this case almost like a fluid (like franklin)?
 
Ok. I understand that I may need to complete more courses before fully understanding why this is the case, but can someone please post some sort of explanation as to why Gauss's law isn't working at this level? I understand the reasoning behind gauss's law, and I just don't see why it should not be applicable to the situation in my original question, which is getting me more confused.
 
On an atomic scale things are moving and wobbling like crazy. Other laws take precedence there: quantum mechanics, to name one. Nevertheless: Gauss' law is holding up just fine at such a level, I would say: around a heavily positive atomic nucleus there is a strong electric field that keeps most of the electrons nicely bound in a close neighborhood -- with a net result of virtually zero field. In a conductor, e.g. a metal lattice, electrons in outer orbits can easily hop over from atom to atom -- and back --, thus smearing out an excess -- or a shortage -- of charge. Concentrating such an excess -- or shortage -- is energetically unfavourable; distributing as far apart as possible is much 'cheaper', so that's what happens.