Gauss's law and surface electrons on a charged conductor, contradiction?

1. Mar 3, 2012

bob900

My textbook (Halliday & Walker) explains that a charged conductor (a solid, of an arbitrary shape) in electrostatic equilibrium will have the electric field inside be 0 and all electrons will be on its surface. It proves this by saying that if the electric field inside was not 0, the free electrons would move and cause a current, and by our equilibrium assumption no such current exists. Then it applies Gauss's law, with an imaginary closed surface S anywhere inside the conductor. Since E inside is 0, flux through S will be 0 too, and so the net charge inside is 0. Therefore, no free electrons will be inside, rather they will all be on the surface.

That's fine, but can't we just as easily apply the same reasoning to these electrons on the surface as well?! Just draw a closed surface S1 around those electrons, but still (infinitesimally) within the conductor. Since those electrons are not moving, the field inside S1 must be 0. And so the flux is too, and so the net charge inside is 0. But we know that S1 encloses a net charge - all those electrons on the surface of the conductor! Seems like a contradiction.

Where is the flaw in this reasoning?

2. Mar 3, 2012

Staff: Mentor

No. Because there is a nonzero net charge inside the Gaussian surface, there is a nonzero net flux of E through that surface. It's the same as the E-field immediately outside the conductor!

In reality, the surface charge layer has some thickness, maybe a few micrometers. As you move into the conductor from the outside, through the layer, the field decreases continuously from whatever value it has outside, to zero. That is, the field is nonzero for a very short distance inside the conductor. At the introductory level, we overlook this sort of thing and pretend that the charge layer has zero thickness and the field drops "instantaneously" to zero as we enter the conductor. This works fine for most practical purposes.

3. Mar 4, 2012

bob900

But if the field is non zero inside the surface charge layer, shouldn't the charged particles within that layer move under the influence of that field?

4. Mar 4, 2012

Anyone?

5. Mar 5, 2012

Hassan2

They don't move because the surface charges can't easily leave the surface of the conductor and accelerate into the surrounding air. In fact there is a strong field due to the nuclei ,holding the charges on the surface. Beneath the surface, the field due to the total charges, pushes the charges toward the surface. On the other hand, the surface charges push the charges back ( because they have the same sign). Since the surface charges cant move outward in the direction normal to the surface, the inner charges come to the the equilibrium, and stay there with a distance from surface.

Last edited: Mar 5, 2012
6. Mar 5, 2012

chrisbaird

Each charge creates its own electric field, whether its moving or not, which exerts a force on the other charges. So a bunch of similar charges, say electrons, will repel each other and spread out in all directions until they meet a barrier, such as the surface. Just below the surface of the conductor, it becomes a balance between the force of the neighboring electrons pushing it ever outwards, and the surface force pulling it in. The balance point is not exactly at the surface, but slightly within it, and location of the electrons depends on the material's surface effect strength (the conductor's "work function") and the total charge/geometry of the objects. The work function really requires a quantum description. The electron becomes delocalized and is bound to the conductor as a whole. Removing an electron from the surface required breaking this bond. If you are interested, google photoelectric effect.

In the textbooks and classroom, though, we just assume for simplicity that the charges are exactly at the surface, that there is no fields anywhere in the object (even just below the surface), and the exact surface is an impenetrable barrier just because it makes these simplifications work out.

7. Mar 6, 2012

bob900

Since we know that the net electric force on the electron is not 0 (as per jtbell's post above), and since the electron is not moving, there must be some other (i.e. non electric) kind of force balancing it out so that the net total force on the electron is 0. From your description it seems like that balancing force is the 'surface force'.

What is its nature? Since it's not electric, it can not be the simple attractive force between the electron and the nuclei of the conductor's atoms. So then where does it ultimately come from?

8. Mar 6, 2012

Per Oni

Hi bob. As far as I am concerned you are quite correct in asking these questions. Take for example jtbell’s few micro meter thick layer. Say such a depth contains roughly 10 000 electronic charges one really has to query the correctness of that statement. I don’t have a “go” at jtbell here because you will find similar statements in all sorts of text books. Imagine the push of all those electrons against the charges which are right at the surface! Add to that that pull of the opposite plate (as in a capacitor).

To say they are not moving because of quantum mechanics is a bit of a conversation stopper, all this can be explained in a classical way.

To realise what happens you must first calculate how many extra charges are flowing to the surface of a capacitor. Then you work out how many there are per unit surface area. Of course you have to start with a reasonable practical situation so that you don’t get spontaneous discharges. You will find then that there are extremely few extra charges in a sea of the normally present charges. So where the notion of “layer” of charges comes from I never understood.

Now you probably still ask: why does this sort of lonely (say) electron not travel to the opposite +ve plate? The answer is that if the electron travels a short distance it will immediately be drawn back due to its mirror image in the –ve plate.

Another much more satisfying answer is to realise that there are not really a few single extra charges which can be pointed out as such. In my view all surface charges are being pulled towards the opposite plate, and therefore each surface charge supplies only a tiny amount of flux in the direction of the opposite plate and most of the flux goes (as before) to the original plate.

9. Mar 6, 2012

Staff: Mentor

You're right about the density of "excess" electrons on the surface. Consider a spherical conductor with radius 0.1 m, charged so it has a potential of 10000 V. I think this is in the ballpark of what you might get with a simple van de Graaff generator. This sphere has a capacitance of $4 \pi \epsilon_0 r$ which is about 10-11 F, so the charge is about 10-7 C, or about 1012 electrons. The surface area is about 0.1 m2, which means there are about 1013 electrons/m2. The spacing between electrons is therefore around 10-7 to 10-6 m.

The spacing between metal atoms is on the order of 10-10 m, so the spacing between "excess" electrons is 1000 to 10000 times the atomic spacing, and they can all be easily accommodated on the outermost layer of atoms without "bumping elbows."

The Wikipedia page about van de Graaff generators claims that for a sphere 0.3 m in diameter, the breakdown voltage in air is about 450000 V. That increases the number of excess electrons by a factor of about 50, and decreases the spacing between them by about a factor of 7, that is, to about 150-1500 times the atomic spacing. Still plenty of elbow room!

Last edited: Mar 7, 2012
10. Mar 6, 2012

bob900

I'm also wondering how it is proven that a (hollow) conductor redistributes charge in response to an external electric field, such that E is 0 everywhere inside.

Just saying "well, if E inside is not 0, the free electrons would be moving and there would be a perpetual current" is not enough of a proof. Maybe when the external field is applied, all the free electrons move (close) to the surface (so there is an initial current), but then they are held there by some combination of forces and cause a non zero E field in the interior of the hollow conductor. And this E produces no current because there are no electrons there - they are all at the surface!

11. Mar 6, 2012

Hassan2

In case of a solid conductor, the extra electrons are at the surface, but the free electrons ( sea of electrons) are everywhere. A nonzero field causes a current then.

12. Mar 6, 2012

bob900

But is it not possible for the nonzero E field to move all the electrons (extra AND free) to, say, one side of the conductor, with the end result being that there will be a net E at points inside the conductor, but yet no movement of charge will occur, because there will be no free electrons left to move.

13. Mar 7, 2012

Hassan2

These free electrons are not really fee. They can freely move from one atom to another only if they are replaced by other electrons (essentially the extra electrons). They can't evacuate the interior and leave a bunch of positive ions.

During charge redistribution, there are free electrons to replace them. Now this question comes to mind:

Isn't it possible to have a nonzero field inside the conductor but there are no electrons to replace the free electron to cause a current?

14. Mar 7, 2012

Per Oni

I hope we are talking about conductors and not isolators. In conductors there are gazillions of free mobile charges. However it could be that I misunderstand your question.

Also: it really is true that there can be no (macroscopic) electrical field inside a (edit: hollow) conductor. I’ve been looking for some articles on the web but can’t find a good one.

Thanks @ jtbell.

15. Mar 7, 2012

Hassan2

Thanks Per Oni,

It was mentioned by jtbell that in reality the surface charge has some thickness, and the field within the thin layer beneath the surface is non-zero. Although we the object is a conductor, we have a nonzero field but no current.

Another extreme case is when the conductor is charged heavily, one extra electron per atom!. In this case, I guess the charges occupy the interior of the conductor too.

16. Mar 7, 2012

Staff: Mentor

I've changed my mind about that. See post #9 in this thread.

17. Mar 7, 2012

bob900

I wonder how that can be proven rigorously.

18. Mar 8, 2012

Hassan2

I read post #9. Thanks. But that was just one case and the voltage breakdown is due to air. Now imagine a thin wire, with 1 mm diameter, and length 2 meters in a field of 1000,000 V/m ( in vacuum). The wire is straight and along the field lines. Charges accumulate on both ends. If we assume that all charges go to the surface of the the ends, we get the density of 442 electrons per square nanometer, which, I doubt it is possible (The field strength at a distance of 1 nanometer from an electron is in larger than 1010 V/m) .

One cause of our confusion is that electricity has a quantum nature but in classical electromagnetics, it is treated as continuum. Maxwell equations describe the filed at any pint as an average over a differential volume around the point containing large numbers of electrons. They l don't describe the field in atomic level.