Electric field inside a conductor?

1. Dec 11, 2009

fluidistic

In the case of a finite conductor rod moving at a constant velocity perpendicularly to a uniform magnetic field, E. Purcell says that the free charges in the rod suffer a force such that they move to an extremity of the rod. When in the final state, he says (at least this is what I understand) that the force exerted on the charges, f is worth -qE where E is the electric field resulting of the polarization of the rod. I think I can understand this thanks to Lorentz force.
But... it means that there's an electric field not null inside a conductor (also outside it). I have had previously difficulties in grasping the fact that inside a charged conductor sphere, the electric field is null.
I realize that there are 2 different cases, but I think I could compare. As an example, in one case there is an excess of negative charges (over the conductor sphere), but the electric field inside the sphere is null.
In the other case, the rod (one could also take a sphere. Let's take a sphere!) is globally neutral, but on one side there's an excess of electrons and on the other side there's an excess of positive ions. However here the electric field inside the sphere is not null? Or is it? In the case of the rod it is NOT NULL. It is also not null outside it.

I have some doubts. I still don't understand well why is the electric field inside a conductor sphere null it seems.

2. Dec 11, 2009

Redbelly98

Staff Emeritus
Actually, there is no net force acting on charges inside an ideal conductor*. So if a conductor moves through a B-field, the charges inside the conductor are also moving through the B-field, and there must be an E-field in order to get zero net force.

[*] Since charges are free to move within an ideal conductor, if there were a net force then the charges would be moving around within the conductor. But if we require that the charges are in an equilibrium configuration, i.e. not moving w.r.t. the conductor, there can be no net force.

Hope that helps.

3. Dec 11, 2009

SystemTheory

For a historical and empirical view look up Faraday Cage or Faraday's Pail.

Let the conductor be isolated. The net charge on the surface is neutral. Now bring in a source that generates an E field external to the conductor. The free charge in the conductor moves about in a tiny tiny tiny fraction of a second to produce a charge distribution on its surface. This rapid process creates an internal E field exactly opposite to the external E field. Inside the conductor the two E fields cancel. There is no net E field inside the conductor and therefore, as Redbelly 98 points out, a test charge placed inside the conductor (if such a thing were possible) would not experience electrostatic force.

The E field outside the conductor does not cancel. It all happens extremely rapidly and depends on the "magical" properties of the free charge carriers in the conduction band of a good conductor material.

4. Dec 11, 2009

fluidistic

I'm very sorry SystemTheory, I realize I didn't write correctly my first sentence "magnetic field, E", It should read "magnetic field, B". So instead of the electric field, consider a magnetic field.
But at least you've explained to me something I wasn't aware of, and it has been useful. So thank you very much!

To RedBelly, I think I understood what you wrote. I realize that in equilibrium (so after a certain time), there's no net force on the charge inside the conductor rod and they're not moving with respect to the rod. However according to Purcell, there is an electric field due to the distribution of the charges inside the rod. I think there are 2 forces on them that cancel out, the Lorentz's force since the rod is moving with respect to the magnetic field and the force due to the induced electric field inside the conductor.
But I don't understand something. I've read that inside conductors, the electric field is always null. Here it is not the case. How can it be so? I'm extremely confused on this.

I took the example of a conductor charged sphere. The electric field inside it is 0. But if I take a neutral conductor sphere and I put it in a magnetic field B with a constant velocity $$\vec v$$ perpendicularly to the magnetic field, the charges will move until reaching an equilibrium in which, I'm not sure whether or not, there's an electric field inside it. (If you replace the sphere by the rod, you have exactly the same case as above, and in the case of the rod, the electric field inside the rod is NOT NULL. This is precisely what I don't understand. The why of this non nullity while in the case of the sphere it seems it's null, despite the charge distribution, but I'm unsure).

5. Dec 12, 2009

Redbelly98

Staff Emeritus
Yes, that's the idea.
If you are using Purcell's Electricity and Magnetism textbook, carefully read section 3.2, Conductors in the Electrostatic Field:
For any conductor, spherical or otherwise, if there are other than electrostatic forces acting on the charge within the conductor, then there will be an electric field present to offset the other force(s).

6. Dec 12, 2009

Bob S

You might be talking about a homopolar generator, also known as a Faraday disk. See
http://en.wikipedia.org/wiki/Homopolar_generator
Sir Mark Oliphant (Australian National University, Canberra) built a large one that stored about 500 MJ of energy in the 1950's.
Bob S

7. Dec 12, 2009

fluidistic

This makes sense. I think I get it.
I've reread Purcell's book (yes RedBelly, the one you provided the name). Indeed, if the reference frame is the static magnetic field, there is an electric field inside the conductor. This is due to the charge distribution induced by Lorentz force, more precisely by the constant velocity of the rod. While, if the reference frame is over the moving rod, there is no external magnetic field anymore, but instead an electric field that can explain the motion of the charges in the rod. However in this frame of reference, there is no electric field inside the conductor rod because the one of the distribution of charges cancels out the external one.