Paradox in electrostatics in relation to Gaussian surfaces? I have encountered something that is very confusing. My problem is this. I am assuming two infinite cubical Gaussian surfaces sharing a common side. One of the cubes contains a charge $q_1$ at a finite distance from the common surface and the second cube  contains a charge $q_2$ at a finite distance from the common surface. 
Now both the charges contribute flux to their respective Gaussian surfaces only at the common surface as the electric field at infinity goes to zero. So, the flux of first cube should be negative to that of second cube, which implies $q_1=-q_2$ which is certainly not true for all cases. Where is my argument flawed?
 A: You have two distinct errors.  One is claiming that because the electric field goes to zero at infinity so does the flux.  The flux is the integral of the electric field over the surface.  The electric field goes down as $\frac 1{r^2}$, but the area goes up as $r^2$, s the flux constant.  Ask Gauss about this.  The second is to claim that because (from the first erroneous argument, but not supported by it) the flux though the five non-common sides is zero, the flux through the common side reflects the charge in each cube.
A: Your problem is the assumption "both the charges contribute flux to their respective Gaussian surfaces only at the common surface."
When you take your surface out to infinity you are also increasing its area, so even though the electric field goes to zero, the integral of the electric field over your surface will be non-zero.
In fact, the total flux through your Gaussian cubes will be constant and equal to $q/\epsilon_0$ no matter the size of the cubes, which is just Gauss' Law.
Your instinct about the common surface is correct: it contributes equal and opposite fluxes to each of your integrals.  However, the other (unshared) surfaces will have a non-zero contribution, and make up for the difference in the magnitude of the charges.
