If the electric flux entering and leaving an enclosed surface...

According to Gauss theorem "the net electric flux through any closed surface is equal to the net charge inside the surface divided by $\varepsilon_0{ }^{\prime \prime}$. Therefore, $\quad \phi \propto \frac{q}{\varepsilon_0}$ Let $-q_1$ be the charge, due to which flux $\phi_1$ is entering the surface, $\phi_1=\frac{-q_1}{\varepsilon_0} -q_1=\varepsilon_0 \phi_1$ Let $+q_2$ be the charge, due to which flux $\phi_2$ is entering the surface. $\phi_2=\frac{q_2}{\varepsilon_0} or \quad q_2=\varepsilon_0 \phi_2$ So, electric charge inside the surface $=q_2-q_1$$\begin{aligned} & =\varepsilon_0 \phi_2+\varepsilon_0 \phi_1 \\ & \Rightarrow \quad=\varepsilon_0\left(\phi_2+\phi_1\right) \end{aligned}$$$