Question

A charge q= -90 nC sits at the center of a cube of side length L = 1.15369248 m. Find a) the total flux of the electric field through the cubical surface. b) the flux through each of its faces, and c_ the flux through the surface of a cube that is twice as large, but is not centered on the charge, q.

Answer 1

Using Gauss's law .

since
q is placed at the center of the cube,

the total flux of the electric field through the cubical surface is Flux = q / $\epsilon$ o [ where q =  -90 nC and $\epsilon$ 0= 8.854 × 10^-12 farads per meter ]
b)

the flux through each of its face is:
Flux = (1/6) * q / $\epsilon$ o [ as cube has six faces]

c) the flux through the surface of a cube that is twice as large, but is not centered on the charge will be same as that of case a).. because the fluc depends on the charge value on upon the size