# Contradictory surface charge density relations

I've seen in many places the relation

$$\sigma = - \epsilon_0 \frac{\partial V}{\partial n}$$

Where $\sigma$ is the surface charge density, $V$ is the electric potential, and $n$ is the direction normal to the surface. For a plane, $n = x$, so this seems to suggest

$$\frac{\partial V}{\partial n} = E_x = -\frac{\sigma}{\epsilon_0}$$

But this seems to contradict the result from Gauss's law using a pillbox as the Gaussian surface:

$$\Phi=2EA= \frac{\sigma A}{\epsilon_0} \rightarrow E=\frac{\sigma}{2 \epsilon_0}$$

What gives? I've written my conclusion as an answer below, but is there a better way of viewing this apparent contradiction?

$\Phi=2EA=\frac{\sigma A}{\epsilon_0}$.

That $2$ should not be there if we are talking about a surface of a conductor that has thickness. The factor $2$ will come when we talk about very thin conducting plates for which the thickness can be ignored, eg: capacitor plates.

However, the $\sigma's$ in the two cases are different. But Gauss's law is not wrong.
Consider a very thin capacitor plate. We can make a Gaussian pill box enclosing just one surface of the plate, one surface is just outside the plate and the other is inside the plate. Inside the plate electric field is zero. Hence,$$\Phi=EA=\frac{\sigma A}{\epsilon_0}$$or,$$E=\frac{\sigma}{\epsilon_0}$$Now we make a Gaussian pill box enclosing both the surfaces of the plate. Now, $$\Phi'=2E'A=\frac{(2\sigma)A}{\epsilon_0}=\frac{\sigma' A}{\epsilon_0}$$or,$$E'=\frac{\sigma'}{2\epsilon_0}$$ Notice that, $E$ and $E'$ are equal as they must be. In the latter, the two areas of the capacitor plate are treated as one single area hence the charge density $\sigma$ must be doubled.
So, its the charge density that we talk about differently in the two cases. I think the confusion has arisen since both are denoted as $\sigma$.
Whatever charge you enclose with your Gaussian surface, it will give the correct result. There is nothing wrong with Gauss's law. The electric field will be obtained correctly.

My conclusion is that the relation from gauss's law, used for a thin sheet, is double-counting the surface charge density. Really, if you want to consider a realistic conductor with some finite-thickness, there is going to be a surface charge on either side of the conductor. Hence, if $\sigma$ is the surface charge on just one surface (identical to the charge on the opposing surface), the total charge enclosed in the Gaussian pillbox would be

$$\Phi = \frac{q_{tot}}{\epsilon_0} = \frac{2 \sigma A}{\epsilon_0}$$

And hence we would get the relation

$$E = \frac{\sigma}{\epsilon_0}$$

From Gauss's law as well.

• Your answer has the word conductor in it, your question does not have the word conductor in it. If it is a conductor with some finite thickness you can place the pillbox to enclose just one surface of the conductor. – Timaeus Sep 5 '15 at 21:50
• Yes, I agree, but this would still change the definition of surface charge from the one I quoted in the common Gauss's law result I quoted above, which is what I was getting at. – aquirdturtle Sep 5 '15 at 22:29
• All I'm saying is that in the common Gauss's law derivation, "surface charge density" has a different meaning than in other cases that you analyze, and that because of this the Gauss's law relationship can be misleading. – aquirdturtle Sep 5 '15 at 22:43
• Whether you want to say this is because the Gauss's law surface is the "only" surface in the Gaussian surface or because the operational definition in the Gauss's law is the surface charge on "both" surfaces seems to be a matter of semantics. – aquirdturtle Sep 5 '15 at 22:43
• I wrote an answer with a corrected version of the law you see everywhere. Its just Gauss's Law. It's always just Gauss's law. That's where it all comes from. – Timaeus Sep 5 '15 at 23:34

Your answer has the word conductor in it, your question does not have the word conductor in it. If it is a conductor with some finite thickness you can place the pillbox to enclose just one surface of the conductor.

Yes, I agree, but this would still change the definition of surface charge from the one I quoted in the common Gauss's law result I quoted above, which is what I was getting at.

The definition of surface charge is a charge per area. If you want to make a surface so thin that you cannot slice it with a Gaussian surface then there are fields on both sides and hence charge on both sides. If it has thickness then you can get the surface charge on just one side by using Gauss's Law to enclose just one side. Something you did not do. I would think you understood except some books (usually more mathematical ones) do start with fields and then define charge from derivatives of fields. They do that to avoid stating exactly which space the charge lives in. I.e. if the fields live in H curl or in H div or in the intersection then you don't have to define what space the charge distribution lives in. In case your books are like that I want to be clear that surface charge already has a definition.

All I'm saying is that in the common Gauss's law derivation, "surface charge density" has a different meaning than in other cases that you analyze, and that because of this the Gauss's law relationship can be misleading.

Whether you want to say this is because the Gauss's law surface is the "only" surface in the Gaussian surface or because the operational definition in the Gauss's law is the surface charge on "both" surfaces seems to be a matter of semantics.

Let's be clear. 1) Gauss's Law is correct. 2) Gauss's Law only tells you the charge enclosed inside a volume of your choosing 3) It is up to you to place the Gaussian surfaces around an actual physical surface to learn anything about that physical surface. When you have two physical surfaces then each produces an electric field so the field is twice as strong because it is due to both. You can get this by superposition. This really happens in a conductor. The field is zero on the inside of the conductor because of the surface charge on both sides of the conductor contribute 4) A Gaussian surface isn't inside a Gaussian surface. 5) Gauss's Law isn't misleading it relates charge enclosed to electric flux 6) When you see an electric field outside a conductor of finite thickness there really is half as much charge on that one surface as you might expect because you really can fit a Gaussian surface inside the body of the conductor. And because the field is half of it due to other charges on the other parts of the surface. They have to be strong enough to produce an equal and opposite field just on the inside so just on the outside they produce an equal and equal field. This is a real effect, not a semantics issue.

Edit If it helps, you can think of the first rule $\sigma = \epsilon_0 \partial V/\partial n$ as a special case. The special case when the electric field right at the surface is only due to the surface charge right there. For a conductor the surface charge right there is only responsible for exactly half the field right there (on the one side of it).

Gauss's Law is always correct. And it is not misleading, you need to use it to learn what it is saying. The equation $\sigma = \epsilon_0 \partial V/\partial n$ is misleading because it only applies sometimes and gives a false sense of generality. It would make you think there is charge in between the plates of a parallel plate capacitor just because there is an electric field there. No one thinks that there is surface charge just because there is an electric field.

You can have some field on two sides of a surface, and then jump indicates some surface charge. This would be the proper version of the $\sigma = \epsilon_0 \partial V/\partial n$ rule.

So the correct rule is $\sigma = \epsilon_0\left( \hat n_1\cdot \vec E_1+\hat n_2\cdot \vec E_2\right)$ where you evaluate on both sides of the surface and the unit normal is always outwards. The jump in the normal component is proportional to the surface charge. Just like Gauss says. And this resolves the senseless sign issue in your version.

• You are misunderstanding what I said was misleading as well as what I said was a semantics issue. My point was that the use of $\sigma$ in this common result from Gauss's law can be misleading. Also, "If you want to make a surface so thin... then there are fields on both sides and hence charge on both sides." -> not necessarily. Take a thick grounded conductor, and put a point charge on one side. one side will gain surface charge and field, and the other won't. This holds in the limit of the conductor getting thinner. – aquirdturtle Sep 6 '15 at 0:15
• The most general form of the rule in terms of potential is $\sigma = - \epsilon_0 (\frac{dV}{dn}|_{above} - \frac{dV}{dn}|_{below})$, which I am well aware of. I also never said that Gauss's law was wrong, I'm just saying you need to be careful with it. – aquirdturtle Sep 6 '15 at 0:20
• @aquirdturtle The first formula you listed is just plain wrong. Gauss's Law is correct and it gives a correct version of the rule you wrote because your rule was wrong. It doesn't matter how common it is, it is wrong. The version I give clearly talks about the charge between two Gaussian surfaces. Which is what you want and should talk about. I might not be following you but I'm not wrong. And a super thick conductor still has charge on the other side. Even if you filled half of space with a conductor there can still be charge at infinity that produces a nonzero field. – Timaeus Sep 6 '15 at 0:21
• The first result is correct for charge at the surface of a non-limit-thin conductor because $\frac{dV}{dn}=0$ inside, which is the case the formula is used to describe. A super-thick grounded conductor has no charge on the other side; there is no reason for there to be. Even if the conductor wasn't grounded, if the plane is infinite, the finite amount of charge will spread out over the infinite surface so the field would be infinitesimal. – aquirdturtle Sep 6 '15 at 0:25
• @aquirdturtle Your version is still wrong, it has some made up orientation. Have the normal go out on both sides and get rid of the minus. You don't have to be careful with Gauss's Law you just have to use it correctly. You put the Gaussian surfaces around the thing you want to find the charge in. Your question didn't mention conductors and had an outright wrong equation for the first equation. – Timaeus Sep 6 '15 at 0:25