Q. 24.1( 9 Votes )

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.

Answer :


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



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