I What's the correct argument for why E=0 inside a conductor?

[Loro]

Hi, I thought that I understood why, once the free charges stop moving, $E=0$ inside a conductor, but I really don't. Can someone please help me out?

I've heard the following arguments, but I don't think I understand any of them:

1. I don't think $q=0$ implies $\vec{E}=0$. I understand that charges end up on the surface, because they experience forces, following from each other's electric fields. Therefore they distribute themselves on the surface of the conductor, such that the electric field lines are perpendicular to the surface of the conductor. Thus, I agree that $q=0$ inside the conductor, but it doesn't imply that $\vec{E}=0$ inside. $\vec{E}$ could be non-zero, but if there are no charges inside, there's nothing to experience a force, so there is equilibrium nevertheless.
2. I don't think the Gauss's law proves anything either. In the integral form $q=0$ implies that $\oint\vec{E}\cdot \text{d}\vec{A}=0$. But it doesn't mean that $\vec{E}=0$. $\vec{E}$ could be non-uniform, so in general there's no problem with having a vanishing flux, but non-vanishing $\vec{E}$. In the differential form $q=0$ implies $\vec{\nabla}\cdot\vec{E}=0$, but we could have a non-zero field with vanishing divergence, there's no problem with that.
3. If that matters, I don't think $\vec{E}=0$ implies $q=0$ either. Even if $\vec{E}=0$ inside a conductor, this itself doesn't imply to me that the free charge has to be vanishing inside. People say that charges won't stop moving until $\vec{E}=0$. But suppose we have a conducting perfect solid sphere with the free charge already distributed on the surface. It can be shown from the Coulomb's law, that $\vec{E}=0$ inside. But that means that if I put a point charge of the same sign as the surface charges, in the very center of the sphere, it won't feel any force. Also, the free charges on the surface would still remain in equilibrium. But now $\vec{E}\neq 0$ inside the sphere, while we still have equilibrium.

So am I wrong in any of the above lines of thought? Or is there another argument that would prove that $\vec{E}=0$? Or do we need any extra assumptions? I'll appreciate any help.

 

[atyy]

http://farside.ph.utexas.edu/teaching/302l/lectures/node28.html

 

[Loro]

They say: "But, the electric field-strength inside a conductor must be zero, since the charges are free to move through the conductor, and will, thus, continue to move until no field remains." but I think I refuted that in my points (1) and (3).

 

[atyy]

Formally, the surface of the conductor is an equipotential surface (since all the charge ends up on not moving on the surface). This plus Gauss's law will show that there is no electric field in the conductor. See 5.3 and 5.4.1 of http://web.mit.edu/sahughes/www/8.022/lec05.pdf.

 

[hilbert2]

You need the condition of the charge distribution being static to show that there's no electric field in a conductor. It also applies to some non-static situations, for instance an electromagnetic wave with a frequency small compared to the speed of redistribution of the charges in the conductor can not penetrate a metal object.

 

[Javier Lopez]

You need the condition of the charge distribution being static to show that there's no electric field in a conductor

Not sure: as higher is the frequency the skin depth is smaller. Slow changing field pass better.
In order to shield waves you should use a superconductopr or plasma.
Do the following test: place the mobile phone inside the microwave oven, close and call to it, but do not switch on the oven!

 

[Dale]

I thought that I understood why, once the free charges stop moving, E=0 inside a conductor, but I really don't

Inside a conductor $J=\sigma E$ where $\sigma$ is the conductivity. Since by assumption the charges are not moving $J=0$ so therefore $E=0$

I don't think $\vec{E}=0$ implies q=0 either

This clearly is implied. If $E=0$ then $\nabla \cdot E = 0 = \rho$

 

[Loro]

Thank you for the replies

Formally, the surface of the conductor is an equipotential surface (...)

I'll think about the uniqueness theorem. I didn't think of that before.

You need the condition of the charge distribution being static (...)

I know. My arguments (1)-(3) all apply to the static case though.

Inside a conductor $J=\sigma E$ (...)

Ok but that's again saying that since charges don't move, there must be no force, which falls into my argument (1).

This clearly is implied. If $E=0$ then $\nabla\cdot E=0=\rho$

That's actually true, thanks. (I'll bold it for anyone else reading the whole thread, because I can't edit the OP) I didn't state the argument (3) properly. What I was trying to say is that charges could stop moving, even though $\vec{E}\neq 0$ as in the example that I described in (3).

 

[Dale]

Ok but that's again saying that since charges don't move, there must be no force, which falls into my argument (1).

Your argument 1 is not relevant. For a conductor $J=\sigma E$. This is not a general fact for all materials, but it is the defining fact of a conductor. If you have a conductor and you have $J=0$ then you have $E=0$ completely irrespective of your argument 1. There is no wiggle room.

I will edit that in the OP.

Please don’t. Editing after you have received responses leads to confusion

 

[Loro]

Your argument 1 is not relevant. (...)

I don't agree. Let's look at the derivation of the microscopic Ohm's law, e.g. here in Sec. 7.5: http://web.mit.edu/sahughes/www/8.022/lec07.pdf
E.g. the whole argument doesn't apply if the are no free charges in the bulk of the material in the first place.

 

[Dale]

the whole argument doesn't apply if the are no free charges

There is no such thing as a conductor with no free charges. If there were no free charges then as you say you could have E without J, but then you would have $J\ne\sigma E$ and thus the material would by definition not be a conductor.

Your argument is not wrong for general materials, but it does not apply specifically for conductors where by definition $J=\sigma E$. I.e. your argument implicitly violates the premise of the problem which is why I said it was irrelevant rather than wrong. It is correct logic that simply does not apply to the given scenario.

What I was trying to say is that charges could stop moving, even though E≠0

Not in a conductor.

Look, my argument here is irrefutable: for a conductor $J=\sigma E$ and for electrostatics $J=0$ therefore for an electrostatic conductor $E=J/\sigma=0$. Since $E=0$ inside the conductor then $0 = \nabla \cdot E = \rho$. Given those two premises (conductor and electrostatic) the rest ($E=0$ and $\rho=0$) follows inevitably. Any counterexample must violate one of the two assumptions.

 

[Loro]

There is no such thing as a conductor with no free charges.

Oh, so am I forgetting that only a small fraction of the free electrons (in case of a negatively charged conductor) end up on the surface, but most of them are still in the bulk of the material, but distributed, such that the net charge of the bulk is zero, and the net charge of the surface is negative?

Any counterexample must violate one of the two assumptions.

But in the argument (3) I give such an counterexample. In it, the central charge is stationary, so $J=0$ but $E\neq 0$. But now that I'm thinking of that it seems to me like an unstable equilibrium, and hence highly unlikely.

 

[Dale]

Oh, so am I forgetting that only a small fraction of the free electrons (in case of a negatively charged conductor) end up on the surface, but most of them are still in the bulk of the material, but distributed, such that the net charge of the bulk is zero, and the net charge of the surface is negative?

Yes, it is a worthwhile exercise to calculate how many free charges are in a typical copper conductor. Even more worthwhile is to calculate the EM stress that would result from removing those charges and compare it to the tensile strength of copper. That comparison would give you a rough estimate of the fraction of charge you could remove before it disintegrated.

But in the argument (3) I give such an counterexample. In it, the central charge is stationary, so J=0 but E≠0. But now that I'm thinking of that it seems to me like an unstable equilibrium, and hence highly unlikely.

Forget the likelihood or the instability. Focus on the two premises (electrostatic and conductor). We know that they are not compatible with your conclusion, so which assumption are you violating? How does remembering that assumption change the situation?

 

[Loro]

Yes, it is a worthwhile exercise to calculate (...)

Ok, got it, thanks - this solves the issue with argument (1).

Forget the likelihood or the instability. (...)

I got it. Putting an extra charge in the center would repell the free electrons in the bulk outwards, such that the surface gets charged more, and the bulk becomes neutral, and $\vec{E}=0$. This refutes the argument (3).

I don't think argument (2) is wrong though.

 

[Dale]

I don't think argument (2) is wrong though.

Well, your argument 2 says that a zero $\rho=0$ does not imply $E=0$ which is true but kind of misses the point that we already proved $E=0$ another way.