When the electric flux associated with closed surface becomes positive, zero or negative ?

If electric flux associated with area $\overrightarrow{\mathrm{S}}$ due to electric field $E$ is $\phi$, then

$\phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{S}}$ $\therefore \phi=\mathrm{EScos} \theta$ $\text { where } \theta \text { is angle between } \overrightarrow{\mathrm{E}} \text { and } \overrightarrow{\mathrm{S}}$ $\text { (i) } \quad \text { If } \overrightarrow{\mathrm{S}} \perp \overrightarrow{\mathrm{E}} \text { means surface is parallel to } \overrightarrow{\mathrm{E}} \text { then, }\theta=90^{\circ}$ $\therefore\text { From equation (1), }$ $\phi=\text { EScos0 }^{\circ}=0$ $\therefore\text{ Flux associated with surface is zero.}$

$(ii)$ If $\theta<90^{\circ}$, then $\cos \theta>0$ (positive) hence flux is positive.

$(iii)$ If $\theta>90^{\circ}$, then $\cos \theta<0$ (negative) hence flux is negative.

All three are shown in figure $(a)$, $(b)$ and $(c) $ respectively.