A cube is placed inside an electric field, $\overrightarrow{\mathrm{E}}=150{\mathrm{y}}^2\hat{\mathrm{j}}\mathrm{V}/\mathrm{m}$. The side of the cube is 0.5 m and is placed in the field as shown in the given figure.  The charge inside the cube is $\mathrm{k} \mathrm{x} {10}^{-11}\mathrm{C}.$ The value of k is__

At bottom surface, electric field is zero as y=0

$\therefore$ Electric flux, ${\mathrm{\varphi }}_1=0$ At top surface, y = 0.5

$\therefore$ Electric flux, ${\mathrm{\varphi }}_2=\mathrm{EA}=\left(150{\mathrm{y}}^2\right)(0.5{)}^2=150\times (0.5{)}^2\times (0.5{)}^2$

$=\frac{150}{4}(0.5{)}^2=\frac{150}{16}$ 

Using Gauss’s law $\mathrm{\varphi }=\frac{{\mathrm{Q}}_{\mathrm{in}}}{{\mathrm{\epsilon }}_0}\Rightarrow \frac{150}{16}=\frac{{\mathrm{Q}}_{\mathrm{in}}}{{\mathrm{\epsilon }}_0}$

$\Rightarrow {\mathrm{Q}}_{\mathrm{in}}=\frac{150}{16}\times 8.85\mathrm{x}{10}^{-12}=8.3\times {10}^{-11}\mathrm{C}\simeq 8\mathrm{x}{10}^{-11}\mathrm{C}$