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 necessarly 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 vec(E) .vec(ds)` represents electric flux over the closed surface, when `oint vec(E) .vec(ds) = 0`, it means the number of flux lines enetering the surface must be equal to number of flux lines leaving it. Further, as `oint vec(E) .vec(ds) = (q)/(in_(0))` where `q` is charge enclosed by the surface. When `oint vec(E) .vec(ds) = 0, q = 0, i.e.,` net charge enclosed by the surface must be zero. Therefore, all other charges must necessarily be outside the surface. This is because charges outside the surface do not contibute to the electric fllux, Choices (c) and (d) are correct.