A charge Q is uniformly distributed over a rod of length ℓ. Consider a hypothetical cube of edge ℓ with the centre of he cub at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.
Given:
Length of rode=edge of cube=l
Portion of rod inside cube=l/2
Total charge =Q
Linear charge density of rod = Q/l of rod=λ
By Gauss’s law, flux of net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ0
…..(i)
Where qin is the net charge enclosed by the surface through which the flux is calculated.
E⃗ =net electric field at the surface
dS⃗ =area of differential surface element
Using gauss law flux through the cube is given by Qin/ϵ0 where
Qin is the charge enclosed by the cube
Charge enclosed by the cube is given by charge density × length of rod inside cube
⇒ 
∴ by (i)
Flux
Therefore flux of electric field through the entire surface of cube is given by Q/2ϵ0
