# 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).

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 $$\vec{E_{net}}=\frac{\sigma}{2\epsilon}-\frac{-\sigma}{2\epsilon}=\frac{\sigma}{\epsilon}$$