When the flux of the electric field through a closed surface is...

According to Gauss's law, the net electric flux through a closed surface is given by $\Phi = \oint \vec{E} \cdot d\vec{A} = \dfrac{q_{enclosed}}{\epsilon_0}$.

Given that the flux $\Phi = 0$, we have $q_{enclosed} = 0$. This implies that the net charge enclosed within the surface must be zero.

The electric field at points on the surface is the resultant field due to both enclosed and external charges. If the net enclosed charge is zero and there are no external charges, the electric field will be zero at all points on the surface. However, if there are external charges, the electric field will not be zero, even though the net flux remains zero. Thus, the electric field may or may not be zero at all points on the surface.