# Using Gauss' Law to find field, simple question

While studying capacitors in my book, the author described how to find the electric field between two flat, metal , charged boards, with uniformely distributed charges along their surface - like in the image shown:

(The border effect was ignored)

The upper board is positively charged and the bottom one negatively charged. It is explained that you should propose a gaussian surface S containing some of the surface of one board (assuming all the charges concentrate on the surface); after applying the law, it finds that the electric field between the boards is $E = \dfrac q{\epsilon_0 A}$, where A is the area of the board.

My problem with this is that the field was found even though there was no account for the charge in the bottom board. I mean, suppose the problem looked actually like this:

The upper half it is exactly the same and so applying Gauss's Law there would yield the same result as before. Except that now the lower board is 99 times more charged than the upper one, so we'd expect the field to be stronger. Also, applying this same surface on the lower board -like it was done on the upper one- would give a stronger field (therefore, 2 different fields on the same point).

What am I missig? My guess is that this law only gives the field due to the charges inside the gaussian surface, and not the actual field - so this kind of process the book used to find the electric field would only work for symmetric configurations. If that is the case, how would one find the resulting electric field of the second example? (assuming $\sigma = \dfrac q{A}$ to be the surface charge density of the boards).

## 3 Answers

Your guess is correct, Gauß's law is $$\oint_A \vec{E}.\mathrm{d}\vec{S}=\frac{q}{\epsilon}$$ where $A$ is the closed surface over which you integrate $\mathrm{d}\vec{S}$. The way to do what you're asking is by superposition. So, you add the fields extracted from the two Gaussian surfaces separately as $\vec{E_1}$ and $\vec{E_2}$ and the resulting field would be $\vec{E_1}+\vec{E_2}$ which would be equal to $\frac{\sigma_1}{\epsilon}+\frac{\sigma_2}{\epsilon}$ in your case, since they are directed along the same line segment.

Even the way you figured out the first field is a little dicy. The field close to an infinitely huge charged surface is not $\frac{\sigma}{\epsilon}$ but $\frac{\sigma}{2\epsilon}$. Why is this the case? Because in case of a metal plate the charge distributes evenly on both surfaces of the plate into $q/2$ and $q/2$, so the charge per unit area is $\frac{q}{2A}$. Now, assuming $\sigma=\frac{q}{A}$, the field is $\frac{\sigma}{2\epsilon}$. The net field is the result of superposition give by $$\vec{E_{net}}=\frac{\sigma}{2\epsilon}-\frac{-\sigma}{2\epsilon}=\frac{\sigma}{\epsilon}$$

• Interesting! In the book it says that the charges on the top concentrate on the lower surface because the charges of the bottom plate creates an electric field that "draws" them downwards (since they move freely). – Dhiego Magalhães Jan 6 '17 at 3:45
• @DhiegoMagalhães: Of course they concentrate, but that is not taken into account in the case of this simple capacitive model. What you are talking about is the question of image charges, but the field existing inside a capacitive setup doesn't require image charges to be considered. – ubuntu_noob Jan 6 '17 at 13:50

What you are missing is that Gauss' law doesn't actually tell you the field in the situation you draw, it tells you the discontinuity in the field as you cross the upper surface charge. So the second situation you draw would have a very different field, but it would have the same discontinuity in the field across the upper surface. To get the actual field there, you have to add a second constraint that is not part of Gauss' law, which is that a conductor must move charges until the field inside is zero. (The tangential component of the surface field also has to be zero, but that's because there can't be a discontinuity in the tangential field.) So the second picture you drew is not consistent with both Gauss' law and the requirement that the interior field be zero in the upper conductor, and that's the source of your confusion.

What are you missing is : Here we are dealing with parallel plate capacitors connected to the same source, so it is not possible to maintain different charges on the plates. You can also get this by assuming that we are applying current to the +ve plate and it is inducing charges on the -ve plate. So, we can get the field by applying Gauss Law on any of the 2 surfaces.