A closed surface in vacuum encloses charges -q and +3q. Another charge -2q lies out side the surface. The total electric flux over the surface is:

$2 q / \epsilon_{o}$

The correct answer is Option (2) → $2 q / \epsilon_{o}$

$\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}$ [Gauss's law]

where,

$\Phi_E$​ = electric flux

$Q_{\text{enc}}$ =​ charge enclosed

$\epsilon_0$ =permittivity of free space

$Q_{\text{enc}} = (-q) + (+3q) = 2q$

$∴\Phi_E =\frac{2q}{\epsilon_0}$