If `oint_(s) E.ds = 0` Over a surface, then

Options

A. The electric field inside the surface and on it is zero.
B. the electric field inside the surface is necessarily uniform
C. the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
D. all charges must necessarily be outside the surface.

Answer

C, D

Explanation

`oint_(S)E.dS=0` represents electric flux over the closed surface. ltBrgt In general `oint_(S)E.dS` means the algebraic sum of number of lux lines entering the surface and number of flux lines leaving the surface. ltBrgt When `oint_(S)E.dS=0`, it means that the number of lux lines entering the surface must be equal to the number of flux lines leaving it.
Now, from gauss' law, we know that `oint_(S)e.dS=0,q=0` i.e., net charge enclosed by the surface must be zero. Thereofre, all other charges must necessarily be outside the surface. this is because charges outside because of the fac t that charges outside the surface do not contribute to the electric flux.