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 lawGiven, Net electric flux, \\\\(\\phi = ({\\phi _2} - {\\phi _1})\\\\) \\\\( \\phi = \\dfrac{q_{in}}{{{\\varepsilon _0}}}\\\\) \\\\(\\Rightarrow q_{in} = {\\varepsilon _0}\\phi \\\\) \\\\(\ herefore q_{in} = ({\\phi _2} - {\\phi _1}){\\varepsilon _0}\\\\) Hence option ‘D’ is correct.