If oint_s E.dS = 0 over a surface, then ______. 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.

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c and d Explanation: We know electric flux is proportional to the number of electric field lines and the term oint_s E.dS represents electric flux over the closed surface. It means oint_s E.dS represents the algebraic sum of the number of flux lines entering the surface and number of flux lines leaving the surface. If oint_s E.dS = 0, this means the number of flux lines entering the surface must be equal to the number of flux lines leaving it. From Gauss’ law, we know oint_sE.dS = q/ϵ_0, here q is the charge enclosed by , the closed surface. If oint_sE.dS = 0 then q = 0, i.e., net charge enclosed by the surface must be zero. Hence all other charges must necessarily be outside the surface. This is because of the fact that charges outside the surface do not contribute to the electric flux.

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