A cube of side l has one corner at the origin of coordinates and extends along the positive `x-,y- and z-axes.` Suppose that the electric field in this region is given by `vec E = (a+by)hatj`. Determine the charge inside the cube (a and b are some constants).

Explanation

The faces adhe, bcgf, cdhg, and abfe will contribute zero flux because the area vector is normal to the electric field for these faces.

Flux through face efgh is
`phi_1 = int vecE * d vec(A) = a(hatj) * l^2 (-hatj) = -at^2`
The field at the face efgh (that lies in the yz- plane , y = 0) is `vec E = a hatj`
and the area vector is `l^2(-hatj)` (direction outward normal). Flux through face abcd is
`phi_2 = (a+bl)hatj*l^2hatj = (al^2 + bl^3)` (for this y = l)
Net flux through the cube is `phi_1 + phi_2 = bl^3`. From Gauss's law,
`phi = (Q_(enclosed))/epsilon_0 or Q_(enclosed) = epsilon_0phi_E = epsilon_0bl^3`.