JEE Main 2023 — Electrostatics Question with Solution

Given:

$\overrightarrow{E}=2x^2\hat{\mathrm{i}}-4y\hat{\mathrm{j}}+6\hat{\mathrm{k}}$

For top and bottom faces, as value of $x=0 \& y=0$, electric field becomes constant and therefore net flux through these two faces will be zero(${\varphi }_z=0$).

Now, $E_x=2x^2$, $E_y=-4y$ and $E_z=6$.

Change in net flux through face at x=1 and at x=0 parallel to y-z plane can written as, ${\varphi }_x=\left(2\times 1^2\right)\times \left(2\times 3\right)-0=12$.

Similarly, net flux through face at y=2 and y=0 parallel to x-z plane can be written as, ${\varphi }_y=-\left(4\times 2\right)\times \left(1\times 3\right)-0=-24$

${\varphi }_{net}=12-24=-12$

Using Gauss's law,

${\varphi }_{net}=\frac{q}{{\epsilon }_0}$

$\Rightarrow \left|q\right|=12{\epsilon }_0$