Question
If $\int_\text{S}\text{E.dS}=0$ over a surface, then:

Answer

  1. The number of flux lines entering the surface must be equal to the number of flux lines leaving it.
  2. All charges must necessarily be outside the surface.
Given,
$\int_\text{s}\text{E.dS}=0$
It means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.
Now, from Gauss' law,
$\int_\text{S}\text{E.dS}=\frac{\text{q}}{\epsilon_0}$
Where q is charge enclosed by the surface.
Now,
$\int_\text{s}\text{E.dS}=0$
q = 0 i.e., net charge enclosed by the surface must be zero.
Hence, all other charges must necessarily be outside the surface.

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