(3, 4) Hint: Use Gauss' law.

Step 1: ${\int }_{\mathbf{S}}\mathbf{E}.\mathrm{d}\mathbf{S}=0$ represents electric flux over the closed surface.

In general, ${\int }_{\mathbf{S}}\mathbf{E}.\mathrm{d}\mathbf{S}$ means the algebraic sum of the number of flux lines entering the surface and the number of flux lines leaving the surface.

When ${\int }_{\mathbf{S}}\mathbf{E}.\mathrm{d}\mathbf{S}=0$, it means that the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Step 2; Now, from Gauss' law, we know that ${\oint }_{\mathbf{S}}\mathbf{E}.\mathrm{d}\mathbf{S}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ where q is charge enclosed by the surface. When ${\int }_{\mathbf{S}}\mathbf{E}.\mathrm{d}\mathbf{S}=0$, q=0, i.e., the net charge enclosed by the surface must be zero. Therefore, all other charges must necessarily be outside the surface. This is because charges outside because of the fact that charges outside the surface do not contribute to the electric flux.