Gaussian surface surrounding only a proton inside a conductor

In summary, the enclosed charge on a proton within a Gaussian surface should be zero, but it is not because of the electric field at any given point on the surface. The electric field is holding the electrons in place, but they can easily hop over to other parts of the surface.
  • #1
RMZ
29
0

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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  • #2
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.
 
  • #3
So should I think about the charge in a conductor in this case almost like a fluid (like franklin)?
 
  • #4
Yes. Consider the conductor as a continuum: homogeneous.
 
  • #5
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.
 
  • #6
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.
 

1. What is a Gaussian surface?

A Gaussian surface is an imaginary surface that is used in Gauss's law to calculate the electric field around a charged object. It is a closed surface that completely surrounds a charged object and helps in simplifying the calculation of the electric field.

2. Why is the Gaussian surface used only for a proton inside a conductor?

The Gaussian surface is used only for a proton inside a conductor because a conductor is an equipotential surface, meaning that the electric field inside the conductor is zero. Therefore, the electric field due to the proton will be the same at every point on the surface, making the calculation easier.

3. Can the Gaussian surface be any shape?

Yes, the Gaussian surface can be any shape as long as it is a closed surface that completely surrounds the charged object. The shape of the surface does not affect the calculation of the electric field as long as it fulfills the necessary conditions.

4. What is the significance of using a Gaussian surface in calculations?

Using a Gaussian surface in calculations simplifies the process of calculating the electric field around a charged object. It allows us to use Gauss's law, which relates the electric field to the charge enclosed within the surface, making the calculations more efficient and accurate.

5. Is the electric field inside the conductor also zero when the Gaussian surface is used?

Yes, the electric field inside the conductor is zero when a Gaussian surface is used, as long as the conductor is an equipotential surface. This is because the electric field due to the enclosed charge is canceled out by the electric field of the conductor, resulting in a net electric field of zero inside the conductor.

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