In the arrangement shown, electric flux through surface 1 is $\mathrm{\varphi }$. Then what will be the electric flux through the closed surface 2 ?

By Guass's law

$$\begin{gathered} {\mathrm{Q}}_1+{\mathrm{Q}}_2={\mathrm{\epsilon }}_0 \mathrm{x} \mathrm{\varphi } ---\left(1\right) \\ \mathrm{and} {\mathrm{Q}}_1+{\mathrm{Q}}_2+8.85 \mathrm{x} {10}^{-6}={\mathrm{\epsilon }}_0 \mathrm{x} {\mathrm{\varphi }}^1---\left(2\right) \\ \therefore \left(2\right)-\left(1\right)\Rightarrow {\mathrm{\epsilon }}_0 \mathrm{x} {\mathrm{\varphi }}^1-{\mathrm{\epsilon }}_0 \mathrm{x} \mathrm{\varphi }=8.85 \mathrm{x} {10}^{-6} \\ \Rightarrow \mathrm{\varphi }=\mathrm{\varphi }+\frac{8.85 \mathrm{x} {10}^{-6}}{8.85 \mathrm{x} {10}^{-12}}=\left(\mathrm{\varphi }+{10}^6\right)\mathrm{v}-\mathrm{m} \end{gathered}$$