A point charge ‘q’ is placed at a distance ‘$\ell$ ’ (on the y-axis) vertically above the semi infinite plane surface which lies in the xz plane as shown in the figure. The magnitude of flux of electric field passing through the surface is $\frac{xq}{48{\in }_0}$  where x is a positive integer. Find x ?

$$\begin{gathered} \varphi ={\varphi }_{OAB}+{\varphi }_{ABCD} \\ =\frac{q}{48{\epsilon }_0}+\frac{q}{48{\epsilon }_0}=\frac{q}{24{\epsilon }_0} \end{gathered}$$

Imagine a cuboid as shown. Flux through left and right face is zero, as electric filed at infinity is zero.

$\begin{array}{l}{\varphi }_{cuboid}=\frac{q}{{\epsilon }_0}\\ \therefore {\varphi }_{face}=\frac{q}{4{\epsilon }_0}\\ \Rightarrow {\varphi }_{OABC}=\frac{q}{16{\epsilon }_0}\end{array}$

Imagine a cube as shown.

$\begin{array}{l}{\varphi }_{cube}=\frac{q}{{\epsilon }_0}\\ \therefore {\varphi }_{face}=\frac{q}{6{\epsilon }_0}\\ \Rightarrow {\varphi }_{OAED}=\frac{q}{24{\epsilon }_0}\\ \Rightarrow {\varphi }_{OAE}=\frac{q}{48{\epsilon }_0}\\ Hence,{\varphi }_{shadedregion}={\varphi }_{OABC}-{\varphi }_{OAE}=\frac{q}{16{\epsilon }_0}-\frac{q}{48{\epsilon }_0}=\frac{2q}{48{\epsilon }_0}\end{array}$