Notice that the flux through the cube does not depend on aaa or ccc. Equivalently, if we were to set b=0b=0, so that the electric field becomes E′→=ai^+cj^E′→=ai^+cj^, then the flux through the cube would be zero. Why?

q_{int} = 0  Ф=0

Explanation:

The flow of the electric field is given by the Gauss equation

             Ф = ∫ E .dA = /ε₀

In this equation we see that the flow is proportional to the charge on the face inside the surface, in this case the charge is zero, therefore the flow must also be zero.

 Another way of looking at it is that the normal vector to the surface has an opposite direction on each of the faces, since the electric field is constant, it can be taken from the integral and the change from the normal to the surface gives zero, so the flow is also zero