Flux through a cube corner

Imagine 8 cubes around a point charge. If you focus only 1 cube, the flux in that area would be $\frac{1}{8}$ of total flux. Since the cube has only 3 sides that electric fields cross, then the flux of 1 side would be $\frac{1}{24}$ of total flux. (The other 3 sides of the cube are parallel to the field, so there is no component that is perpendicular to those sides left.) The answer would be $(\frac{1}{24})(\frac{q}{\epsilon_0}) = (\frac{1}{24})(\frac{96\epsilon_0}{\epsilon_0}) = 4$ .

By Gauss' law , a surface containing the entire charge has total electric flux $q/ \epsilon_0 = 96$ . However, the cube as shown only contains one-eighth of the total flux. Furthermore, each of the three sides that contain non-zero flux has exactly one-third of the total flux of the cube. Therefore, the flux that goes through just one of the three sides is $96 / 24 = 4$ .

You can consider a cube C centered at the origin and with length-side two. The flux is the equal to 96/6 over each of the six faces of such a cube. The area considered is one fourth of of the area of the face C and by simmetry we obtain flux=96/6/4=4