# Gauss Law of a uniformly charged plane

We know that Gauss Law for some volume $V$ with surface $\partial V$ is:

$$\int_{\partial V} \vec E \cdot \vec n dS = \int_V \frac{\rho}{\epsilon_0}dV$$

What it means is that if there is charge inside $V$ then it will generate a field and the total flux of that field through the surface $\partial V$ is proportional with that charge. If there is charge outside $V$ its contribution to the flux will cancel out.

This means that it does not matter if there is charge outside $V$, the only field that is taken into consideration is the one generated by the charge inside. To make this clear, we can write $\vec E = \vec E_{out} + \vec E_{in}$ where $\vec E_{out}$ is the field that does not contribute to the total flux through $\partial V$ and $\vec E_{in}$ is the one generated by the charge inside $V$. Plugging this in Gauss Law, we get:

$$\int_{\partial V} \vec E_{in} \cdot \vec n dS = \int_V \frac{\rho}{\epsilon_0}dV$$

So it does not matter whether there is or there isn't something outside $V$, all we need is the field generated by what is inside it.

Now, let's consider the case of the uniformly charged plane. We have the electric field perpendicular to the plane, and we will consider a box as here. If we apply the same logic as in the beginning, the total flux through that box is given by the field generated by the charge in the box only. So we can ignore the rest of the charged plane, and if we do so, the field that remains is only $\vec E_{in}$. But there is a problem. This resulting field is not perpendicular to what used to be the plane and it is not as easy to integrate anymore in order to get the total flux. In fact, on the curve where the charge distribution intersects the surface, $|\vec E|\rightarrow\infty$ so not a very easy to analyse scenario.

Now in all the cases I have seen, this problem of integration and divergence of the field is not mentioned because the authors use the perpendicular electric field and everything is solved within 3-4 lines. So in order to explain why that is the case, I guessed that, although it is not mentioned, in order to evade the problem with the integration, they add a very specific $\vec E_{out}$ that solved the difficulties of the integration and divergence but does not contribute to the result?

What I'm interested in, regarding this problem is the validity of my reasoning. Am I just overthinking a problem that can be solved with some well placed approximations (height of box very small, surface covered very big) or what actually happens is exactly what I described above but people simply skip that part whenever they solve this problem?

Edit:

I know how to do the computation and also I'm familiar with the way this problems is usually solved. I will reformulate the question based on the feedback received so far.

Consider the plane and take some cylinder as in the link I used above. Now we will take 2 cases.

Case 1. - We consider the infinite plane and the cylinder

Case 2. - We consider only the part of the plane that is inside the cylinder. The charge outside the cylinder does not exist anymore.

For both cases, the right-hand side of Gauss law is the same, that is the total charge inside the volume. However, the field in case 1 is different than the one in case 2.

More to that, and based on the observation in the comments that Gauss law is about the flux and not the field. In case 2, the surface defined by the charge distribution intersects the cylinder surface, thus resulting a curve $\Gamma$. The field around $\Gamma$ goes to $\infty$. If I want to compute the flux in this case the integral wouldn't converge. But this is if I try to compute the flux. On the other hand, the charge is finite. How does this happen?

There is no need to say that for the infinite plane these problems do not appear.

Here is my problem to which I tried to answer and asked if the reasoning is ok. It should be the same result, in both cases, but case 2 does not look good. Where am I wrong?

• I don't think there's a problem. The net fluxes for the two cases (infinite sheet and finite disk withing Gaussian surface) are the same. But the field distributions aren't. And you will not get the correct field distribution for an infinite sheet by considering a finite disk. – garyp Sep 1 '17 at 20:36
• @ZeroTheHero Fixed the linguistic issue. About Gauss law referring to the flux, not the field, I added another part to the question that should be better formulated. – Victor Palea Sep 1 '17 at 22:18
• How are you so sure that the integral will diverge? – Javier Sep 1 '17 at 22:24
• @Javier I started from the Gauss surface and the charge distribution surface intersection curve $\Gamma$. By using $E~q/r$ where $r$ is the distance between some charge $q$ and $\Gamma$, then I get $lim_{r\rightarrow 0}E(r)=\infty$. – Victor Palea Sep 1 '17 at 22:42
• I know the field diverges, but the integral shouldn't. After all, it must be equal to the enclosed charge. – Javier Sep 1 '17 at 23:05

## 2 Answers

Firstly, You can determine from Coulomb's law that the electric field is perpendicular to the infinite sheet. This is true at every point on the sheet. This does not change once you put an imaginary Gaussian surface to capture some of the electric flux in some region of the sheet. The electric field will still point perpendicular to the sheet in the region inside the Gaussian surface.

Let's do the computation. The sheet will be in the $xy$ plane, with the Electric field pointing in $\hat{z}$ Our gaussian surface is the one you have in the link. We have the following $$\int_{\partial V} \vec{E} \cdot \vec{n} dS = \int_{cylindrical \\surface} \vec{E} \cdot \vec{n} dS + \int_{top \\disk} \vec{E}\cdot \vec{n} dS + \int_{bottom \\disk} \vec{E}\cdot \vec{n} dS= \int_V \frac{\rho}{\epsilon}dV = \int_{region\\ on \\ sheet} \sigma dA$$

Now on the cylindrical surface $\vec{n} =\hat{r}$ and on the disks $\vec{n} = \hat{z}$. This means the integral on the cylindrical surface is $0$ not $\infty$ since $\vec{E}= |E|\hat{z}$. All we are left with is the integral over the disks.

As a consequence of this it is the electric flux inside the gaussian surface that contributes.No where have we made any approximations. What allowed this simple result is that the sheet was infinite in the $xy$ plane otherwise we might have had the inhomogeneities you are worrying about.

Addition:

Case 2 changes the problem, although the electric flux inside the Gaussian surface is what contributes, we need the charge outside to keep the electric field constant everywhere otherwise we will not have the cancellations in Coulomb's law that make the electric field point in the same direction everywhere. So therefore when you say, "The charge outside the cylinder does not exist anymore". You have changed the problem, the infinite sheet is no longer infinite and then you do have the problem you are worrying about.

• I added another part to the question that addresses the aspect you mentioned in the last paragraph. – Victor Palea Sep 1 '17 at 22:22
• @VictorPalea , I hope I am making sense. Try to determine the direction of the electric field you will find that you need the charges outside and inside the gaussian surface. The charge outside is needed to keep the calculation exact i.e to make the electric field point in the same direction everywhere and have the same magnitude – Amara Sep 1 '17 at 22:43

I see you remain confused with this, but honestly I really don't get why.

1) The fact that $\vec{E}$ is perpendicular for an infinite sheet has nothing to do with the Gauss Theorem. If the plane is infinite, the electric field must be perpendicular to it, because that's the only way it is indisinguishable from one point to another. If there were something different, that point would be a sign and then every point wouldn't be equivalent to the rest, so that would be a different problem.

Consequently, the field is perpendicular to the plane.

2) Now you apply Gauss law as @Amara says in the other answer. You are not taking into account any outer field. You are only using that $\vec{E}\cdot \hat{n}= 0$ in the wall of the cylinder. You only have contributions in the bottom and the top disks.

In each disk, you have $\int E\cdot dS$. Now you use that $E$ must be constant all along the disk (because the disk is parallel to the surface, and we said that every point in the surface is equivalent). Consequently, $E$ is constant and you have

$E\cdot \iint dS= E\cdot S$, and that's equated to $q/\varepsilon_0$. That's why you have $E=\frac{q}{S \varepsilon_0}\equiv \frac{\sigma}{\varepsilon_0}$.

Is your question why $E$ is constant along that disk (angle included)? As I told you, every point in the sheet creates a perpendicular electric field (only perpendicular) due to symmetry. It doesn't matter if you restrict the area of integration to a single disk: your system is creating a field only perpendicularly. It is NOT equivalent to a system in which you have a charged disk because the field a disk generates is different, so you'd be integrating a different quantity. I don't know if that's what you're asking.