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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.
Solution
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.
EL = Cx = Ca (x = a at left face)
Magnitude of E at right face
ER = 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
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