Help! No net charge inside conductor

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Homework Statement



Prove that there is no charge inside a SOLID conductor, and thus, charges are only on the surface.

Homework Equations



Gauss's Law (Maybe)

The Attempt at a Solution



I have been doing some research on it on some websites.

It mentioned that, "you make a Gaussian surface inside the conductor. And since we know that the electric field inside any conductor is 0 so there is no charge inside the conductor..."
The problem is that, those websites haven't prove that the electric field inside a conductor is perfectly zero ( which I find mind-blowing ) .

Can anybody provide a proof (preferably mathematical) either that the net [itex]\vec{E}[/itex] inside is zero or that there is no charge inside?

Any help is very much appreciated.
 
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Answers and Replies

  • #2
rude man
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The argument is qualitative.

Take an uncharged conducting solid. Implant one electron in it. It can roam anywhere; there is no force on it anywhere.

Put a second electron on the solid. Now both these charges 'see' each other and repel each other. They will move as far away from each other as possible, i.e. to the surface.

Put a third, etc etc electrons on it, same deal. The all repel each other & so will distribute on the surface.

I suppose you could take the Poisson equation div{grad(V)} = -ρ/ε and E = -grad(V) with the appropriate boundary values to show that ρ = 0 everywhere inside the solid, but I wouldn't know how or even if.
 
  • #3
Office_Shredder
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If the conductor is in an electrically steady state, and there is an electric field inside of it, any electrons in the region with an electric field will experience a force and start moving, contradicting the steady state claim
 
  • #4
fluidistic
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What about the special case of a conductor sphere with an "electron" (let's take it as an idealized classical electric charge, be it positive or negative and absolutely quiet) right at the center of the sphere? Clearly, the E field inside the conductor is not 0 and clearly there's no force applied on the "electron". I agree that it's an unstable equilibrium point, any motion of the charge and it goes on the surface, but still.
Now if you start with the assumption that inside any conductor the E field is null (discarding the above case) then using Gauss law is sufficient I think to prove that any charge must lie on the surface of the conductor. Because [itex]\vec \nabla \cdot \vec E =4\pi \rho[/itex] if E equals 0 then rho must equal 0.
A bit more formally:
axioms:
1)The E field inside any conductor is 0.
2)There's an excess of charge somewhere in the conductor.
"Proof": For any region [itex]\Omega[/itex] inside the conductor, [itex]\int _{\partial \Omega } \vec E \cdot \hat n da=0[/itex] because E is worth 0 there according to axiom 1. Using Gauss law under the integral form, it's equivalent to say that [itex]4\pi \int _{\Omega } \rho d^3x =0[/itex] for any region omega inside the conductor. But for this to happen, rho must be 0 for any region omega inside the conductor. Since we assumed that there was an excess of charge somewhere in the conductor (axiom 2), the only remaining possibility is that the excess of charge lies over the surface of the conductor. End of proof.
P.S.:I used units in which k=1 rather than [itex]\frac{1}{4\pi \varepsilon _0}[/itex].
 
  • #5
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As the others have observed, usually we take it for granted that the electric field inside a conductor is very nearly zero. This is based on the empirical knowledge that conductors have the special property that there are a lot of relatively mobile electrons floating around, which will rearrange themselves very quickly to restore electrostatic equilibrium if it is disturbed.

What about the special case of a conductor sphere with an "electron" (let's take it as an idealized classical electric charge, be it positive or negative and absolutely quiet) right at the center of the sphere? Clearly, the E field inside the conductor is not 0 and clearly there's no force applied on the "electron". I agree that it's an unstable equilibrium point, any motion of the charge and it goes on the surface, but still.
So, in this case the net electric force on the central electron may be zero, but it introduces a non-zero electric field that exerts a force on all the other conduction electrons. These electrons will move out towards the surface, inducing a negative charge on the surface. Since there are now fewer electrons near the center, the positive charge from the lattice nuclei are less shielded and they can cancel out the non-zero field induced by your idealized charge. So even if you fixed that charge there mechanically, the conductor would still rearrange itself into equilibrium, if possible.
 
  • #6
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Thank you very much for these great answer!
 

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