A cube has one corner at the origin and the opposite corner at the point (L, L,L)(L, L,L). The sides of the cube are parallel to the coordinate planes. The electric field in and around the cube is given by

Net charge = E• b • L^3.

Explanation:

NB: here, the symbol representation of the flux is "p" = electric Field • Area(dot Product).

So, we will take a look at the flux through -x face, through x face and through -y face, through y face and through - z face and through z face.

(1). Starting from -z and z faces which are the back and front faces of the cube:

Thus, We have that the flux,p = 0 for -z and z.

(2). Recall that we are given that E = =(a+bx)i^+cj^.

Thus, p_-y = (a + bx)i + cj (-j) (L^2)

Where y = 0

p_-y = -cL^2.

Obviously for p_j, we will have cL^2 and y = L

(3). For p_-x = =(a + bx)i + cj (-i) (L^2).

p_-x = -aL^2

Where x = 0.

When x = L and p_x = (a + bL)L^2.

This, adding all together gives Net charge = E • b • L^3.

i think the answer is b

object 2 because all the other ones are wrong