Q. 84.8( 5 Votes )

A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is



A. zero

B. q/ϵ0

C. q/2ϵ0

D. 2q/ϵ0

Answer :

Gauss’ law states the electric flux through an enclosed surface is 1/ϵ0 times the charge enclosed by the surface.


where


𝚽 = electric flux,


E = electric field,


ds = area element,


q is the charged enclosed,


ϵ0 is the electric permittivity of vacuum.


In this case, the surface is not closed, since we are considering the open end of a cylindrical vessel. We construct another identical cylindrical vessel on top of it.



Now, the charge enclosed by two of the surfaces is q.


Hence, the charge enclosed by only of the surfaces will be q/2.


Therefore, the net flux will be half of the total = q/2ϵ0.

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