Question

State Gauss’s law in electrostatics. A cube which each side ‘a’ is kept is an electric field given by `vecE` = C × l. (as is shown in the figure where C is a positive dimensional constant. Find out

(i) The electric flux through the cube, and

(ii) The net charge inside the cube.

Answer 1

Gauss’ law states that the total electric flux through a closed surface that enclosed a charge is equal to

`1/in_0`times the magnitude of the charge enclosed.

Here, ∈₀ is the absolute permittivity of the space

`phi=q/in_0`

q is the total charge enclosed

Also,`phi =oint_s vecE.vecds q/in_0`

Where  `vecE`is the electric field at the area element`vecds.`

 

(ii) Now, the electric field  `vecE=Cxhati` is in x-direction only. So faces with surface normal vector perpendicular to this field would give zero electric flux i.e. ∅ = Eds cos 90° = 0

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 as (x = 2a at right face)

Thus, corresponding fluxes are

`phi_L = E_L *ds =E_L ds cos theta`

                       = `-a C xx a^2 (as theta = 180°)`

`phi^R = E_R *ds = 2a  C ds costheta (theta =0)`

                            `=2aCa^2`

                            `=2a^3C`

Now, net flux through cube is

`=phi_L +phi_R`

`= -a^3 C+2a^3C`

`=a^3CNm^2C^-1`

(b) Net charge through cube

Again, we can use Gauss’s law to find total charge q inside the cube.

We have

`phi = q/in_0  or q =phi in_0`

`q =a^3Cin_0`Coulomb