Give a reason: 'If the net flux associated with a closed surface is zero,then the net charge enclosed by that surface is zero.'

(N/A) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$,where $q_{enclosed}$ is the net charge enclosed by the surface.
If the net flux $\phi = 0$,then $\frac{q_{enclosed}}{\epsilon_0} = 0$,which implies $q_{enclosed} = 0$.
This means that the total charge enclosed by the surface must be zero. The flux through a closed surface is zero if the net charge inside is zero,even if there is an external electric field present,as illustrated by the case of a cylinder in a uniform electric field $\vec{E}$.
For a cylinder placed in a uniform electric field $\vec{E}$ parallel to its axis:
$1$. The flux through the curved surface $3$ is $\phi_3 = 0$ because the area vector is perpendicular to $\vec{E}$ at every point.
$2$. The flux through the flat surface $1$ is $\phi_1 = -ES$ (where $S$ is the cross-sectional area,and the normal is opposite to $\vec{E}$).
$3$. The flux through the flat surface $2$ is $\phi_2 = +ES$ (where the normal is along $\vec{E}$).
The total flux $\phi_{total} = \phi_1 + \phi_2 + \phi_3 = -ES + ES + 0 = 0$. Since the total flux is zero,the net charge enclosed is zero.