b
Hint: Here, we have to apply gauss's law to find electric charge.

Solution:

We know that, electric flux \(\phi_1\) (or electric field lines) entering in a closed

surface is -ve and electric flux \(\phi_2\) (or electric field lines) leaving a closed surface is \(+ve.\)

Hence, net electric flux through the closed surface,

\(\phi=\phi_2-\phi_1\)

Now, according to Gauss' theorem, the net electric flux \(\phi\) passing through a closed surface is equal to the \(1 / \varepsilon_0\) times of the total charge \(q\), inside the surface.

Step1: Apply gauss's law

Given, Net electric flux, \(\phi=\left(\phi_2-\phi_1\right)\)

\(\phi=\frac{ q _{\text {in }}}{\varepsilon_0}\)

\(\Rightarrow q_{\text {in }}=\varepsilon_0 \phi\)

\(\therefore q_{\text {in }}=\left(\phi_2-\phi_1\right) \varepsilon_0\)