# Electric field near the surface of a conductor

1. Jun 28, 2011

### pc2-brazil

Good afternoon,

I was self-studying Electricity (Gauss' Law) and I have a doubt regarding the electric field near the surface of a conductor.
I know that, near the surface of an infinite plate made of a non-conductive material, the electric field can be given by:
$$E=\frac{\sigma}{2\epsilon_0}$$
However, near the surface of a conductive body, the expression becomes:
$$E=\frac{\sigma}{\epsilon_0}$$
which is twice the expression above.
I understand how this can be explained using Gauss' law. But I'm trying to think about it only in terms of Coulomb's law.
If the conductor is an infinite thin blade, then I understand that the charges will be distributed over its entire external surface, so that the configuration will become analogous to that of two non-conductive infinite plates parallel to each other, and the electric field near the surface will be
$$E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2 \epsilon _0}=\frac{\sigma}{\epsilon_0}$$
yielding the expression for E near the surface of a conductor. Also, between the two plates (that is, inside the conductor), E will be zero, which is consistent with Gauss' law.
But what if the conductor is a solid sphere or a solid cube, for example? When I try to think about this using only Coulomb's law, I don't see why the electric field near the surface of the conductor couldn't be given by the first expression above. As the charges are distributed over the entire external surface, if we are very near to the surface, wouldn't the configuration be analogous to that of one infinite non-conductive plate?

2. Jun 28, 2011

### RedX

A nonconducting sphere of surface charge density $\sigma$ has a field just outside the sphere of $$E=\frac{Q}{4 \pi \epsilon_0r^2}=\frac{\sigma 4 \pi r^2}{4 \pi \epsilon_0r^2}=\frac{\sigma}{\epsilon_0}$$, and zero field on the inside.

This is exactly the same as a conducting sphere or ball.

3. Jun 29, 2011

### Philip Wood

This is a good question.

Divide the (positively) charged surface of the conductor into a small 'local' patch, P, and the rest of the surface, R.

At a point, I, just inside the conductor from the centre of P, the field ($\sigma$/2$\epsilon$0) due to P is pointing inwards (away from P), and the field due to R is pointing outwards. Since there is no net field at I under electrostatic conditions (or free electrons would be drifting), we know that at I the fields due to P and R must be equal and opposite.

Now step a tiny distance from I to a point O just outside the conductor from the centre of P. The field at O due to P is equal and opposite to the field at I due to P. But the field due to R, which is relatively remote, is the same at O as at I. It can't have changed its direction, or indeed its magnitude, in so small a distance.

So the field at O, which is the sum of the field due to P and the field due to R, which are BOTH directed outwards is 2 ($\sigma$/2$\epsilon$0).

4. Jul 3, 2011

### Philip Wood

Field just outside a charged conductor without direct appeal to Gauss's law. Here's the same argument spelt out a little more carefully ...

Suppose the conductor's surface has a positive charge spread over it. Consider a point O just outside the surface. Using Coulomb's law, it's easy to show by direct integration that the field strength due to the charge on the nearby surface is σ/2ϵ0, directed normally outwards from the surface. How nearby, you ask? By changing the integration limits we can show that if we are a small distance, d, from the surface, then 99% of σ/2ϵ0 is contributed by a circular patch of surface of radius 100 d (centred on the point on the surface closest to O). This assumes that the surface is effectively plane over this local area.

So far we've considered the field strength, EL, at O due to local surface charge on the conductor. But there will also be a field, ER, at O
arising from 'remote' charges, i.e. charges not on the local patch of surface. These remote charges could be on other conductors (e.g. the opposite plate in a parallel plate capacitor) and/or elsewhere on the same conductor - remember that the surface of a real conductor can't be an infinite two dimensional plane, but must be topologically equivalent to a spherical surface: it can't stay flat indefinitely.

There is a delightfully neat argument which tells us how ER compares in magnitude (and direction) with EL. We consider a point I which is close to O but just inside the conductor. We know from Coulomb's law that the field strength at I due to the local surface patch is -EL, i.e. of magnitude σ/2ϵ0, but directed inwards. But the field due to the remote charges is ER, just the same as at O, because the distance between I and O can be made infinitesimal in comparison to the distances away of the remote charges. We also know that the resultant field at I is zero. If it weren't zero, then it would give the free charges in the conductor a drift velocity, which is ruled out by this being an electrostatic set-up. So we have
ER + (-EL) = 0. So ER = EL. So, outside the conductor, the field strength is ER + EL = 2EL = σ/ϵ0 directed outwards.

5. Jul 3, 2011

### pc2-brazil

Very good explanation. I understand it now.
I understood your previous post too, but this more detailed explanation confirmed what I had understood.

Last edited: Jul 3, 2011