Physics - Electrostatics Question with Solution | TestHub

Given here, electric field, $\overrightarrow{E}=E_{0}x\hat{\mathit{i}}$

Here, the flux passes mainly through surface areas, $ABCD \mathrm{and} EFGH$. As the surfaces $\mathrm{AEFB} \mathrm{and} \mathrm{CGHD}$ are parallel to the electric field, so flux for these surfaces are zero. Again in $EFGH$, $a=0$, thus, electric field is zero.

Hence, the flux only passes through the surface are $ABCD$.

The net flux is ${\varphi }_{\mathrm{net}}={\varphi }_{\mathrm{ABCD}}=E_{0}a·a^2$

Using Gauss's law, $\frac{q_{\mathrm{en}}}{{ϵ}_0}=E_0{a}^3$

$\Rightarrow q_{\mathrm{en}}=E_0{\epsilon }_0{a}^3$

$=4\times {10}^4\times 9\times {10}^{-12}\times 8\times {10}^{-6}$

$=288\times {10}^{-14} \mathrm{C}$

Hence, the value of $Q=288$.