# 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

We know that,

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.

E =net electric field at the surface

dS =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 Rate this question :

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