State Gauss’s law in electrostatics. A cube with each side a is kept in an electric field given by E = Cx (as shown in the figure) where C is a positive dimensional constant. Find out :

(i) The electric flux through the cube (ii) The net charge inside the cube.

Explanation

Gauss’s law states that the total electric flux through a closed surface is equal to `1/e_0` times the magnitude of the charged by enclosed by it.
`phi=e/_0`
Here, `e_0` is the absolute permittivity of the free space and q is the total charge enclosed.

Also `phi= int E.ds =q/E_0`
Where, E is the electric field at the area element dS. Now, the electric field is in X-direction only. So, face with surface normal vector perpendicular to this field would give zero electric flux i.e.,
`phi=E ds cos 90^@=0 ` through it.
So, flux would be across only two surfaces
Magnitude of E at left face ,
`E_L=Cx=Ca`[x=a at left face]
Magnitude of E at right face
`E_R=Cx=C2a=2aC`[x=2a right face]
Thus, corresponding fluxes are
`phi=E_l.ds =E_L ds cos 180^@=-ac times a^2=-a^3c, phi_R=E_R.ds=2ac Dc cos0^@=2aca^2=2a^3C`
(i)Now, net flux through cube `=phi_L+phi_R=-a^3C+2a^3C=a^3C Nm^2C^-1`
(ii) We have, `phi=e/e_0 or e=phi e_0, Q=a^3 C e_0` Coulomb