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(D) According to Gauss's law,the net electric flux $\phi_{net}$ through a closed surface is equal to the total charge $q$ enclosed by the surface divi

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$\varepsilon_0(\phi_2-\phi_1)$

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(D) According to Gauss's law,the net electric flux $\phi_{net}$ through a closed surface is equal to the total charge $q$ enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.$\phi_{net} = \frac{q}{\varepsilon_0}$Here,the flux entering the surface is $\phi_1$ (which is negative) and the flux leaving the surface is $\phi_2$ (which is positive).Therefore,the net flux is $\phi_{net} = \phi_2 - \phi_1$.Substituting this into Gauss's law:$\phi_2 - \phi_1 = \frac{q}{\varepsilon_0}$$q = \varepsilon_0(\phi_2 - \phi_1)$.

If the electric flux entering and leaving an enclosed surface are $\phi_1$ and $\phi_2$ respectively,the electric charge inside the surface will be

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