How does one apply Gauss's Law when dealing with two infinite planes? 
If you have two parallel planes with one density of 1 C/m^2 and the other with -1 C/m^2 density and you need to find the electric field at a point in the middle, how do you use Gauss's Law?

I drew a box enclosing one plane to find the electric field in the midpoint. I understand there's another plane exerting an additional force, but by Gauss's law it should only depend on the enclosed charge. 
Is Gauss's Law being violated here?
 A: you don't have to find it through gauss' law. you can simply calculate the resultant electric field  by vector addition of the electric field of every plane as how you calculate resultant force while many forcers are applied on a particle.
A: No, Gauss law is not at all violated here.

When you considered a single plate you get flux $\phi$ in box ABCD. Total field would be $\phi$ as sum on two sides AB and CD.
Let here flux towards AB is $\frac{\phi}{2}$ and towards CD is  $\frac{\phi}{2}$ as well.
Net flux = $\phi$. _________(1)

When second plane of -ve charge is added.
You again get an electric field. But this time you see that E is towards -ve plate. Flux $\phi$ entering the side CD is equal to flux exiting side AB .
Net flux due to -ve plate = +$\frac{\phi}{2}$ - $\frac{\phi}{2}$ = 0 ._______(2)
On adding (1) and (2)
Total flux = $\phi$ +0 = $\phi$
As Gauss law stated Net flux from ABCD = $\frac{+q}{\epsilon}$
which is independent of -q on other plate as no net flux entered due to it.
A: The electric field due to an infinite plane is constant in the entire space around it so if you have two planes with opposite charge density the electric field between the planes will be zero. And double on the out side of the planes
