Gaussian surface surrounding only a proton inside a conductor

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Homework Help Overview

The discussion revolves around the application of Gauss's law in the context of a Gaussian surface surrounding a single proton within a conductor. The original poster expresses confusion regarding the expected zero net charge when enclosing a proton, despite understanding that conductors should have zero charge at any point.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of applying Gauss's law at subatomic scales and question the validity of classical electrodynamics in this context. Some suggest considering the conductor's charge distribution as fluid-like, while others emphasize the need to account for quantum mechanics at atomic scales.

Discussion Status

The discussion is ongoing, with participants providing insights into the limitations of classical approaches at small scales. There is recognition that Gauss's law may still hold, but its application requires careful consideration of the underlying physics at the atomic level.

Contextual Notes

Participants note that the behavior of charges in conductors is influenced by macroscopic properties and that classical laws may not fully apply at the atomic scale. There is an acknowledgment of the need for further study to grasp these concepts fully.

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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Being a conductor is a macroscopic property and you are talking about a Gaussian surface at subatomic scales, such a small scale that classical electrodynamics are put out of play.
 
So should I think about the charge in a conductor in this case almost like a fluid (like franklin)?
 
Yes. Consider the conductor as a continuum: homogeneous.
 
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.
 

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