Gauss law states that flux due to net electric field on a Gaussian surface is charge enclosed divided by permittivity.
$$\Phi = \oint _ { A } \vec { E } \cdot \mathrm { d } \vec { A } = \frac { Q } { \varepsilon _ { 0 } }$$
Let's take a spherical Gaussian surface with $+q$ charge at centre. We can simply calculate the electric field at the Gaussian surface. Now if we introduce a charge $-q$ outside the surface at a point outside the surface and $r$ away from surface ($r$ is radius of Gaussian surface). The electric field at the surface would be same by Gauss law, but how can it be. Won't the point equidistant from the charges and on the surface experience $0$ electric field?
