JEE Main 2025 — Electrostatics Question with Solution

To calculate the area of the surface through which the electric flux is determined, we use the given electric field and the specified surface orientation. The electric field is expressed as $\overrightarrow{\mathrm{E}}=(2\hat{i}+4\hat{j}+6\hat{k})\times 10^{3}N/C$.

Since the surface is parallel to the $x-z$ plane, its area vector $\overrightarrow{\mathrm{A}}$ is directed along the $y$-axis, making it $\mathrm{A}\hat{\mathrm{j}}$.

The electric flux $\varphi$ through the surface is given by:

$\varphi =\overrightarrow{\mathrm{E}}\cdot \overrightarrow{\mathrm{A}}=(2\hat{i}+4\hat{j}+6\hat{k})\times 10^3\cdot \mathrm{A}\hat{\mathrm{j}}$

Calculating the dot product, the only component contributing to the flux is the $y$-component:

$\varphi =(4\times 10^3)\mathrm{A}$

Given that the electric flux $\varphi$ is $6.0{\mathrm{Nm}}^2/\mathrm{C}$, we can equate and solve for $\mathrm{A}$:

$6=4\times 10^3\mathrm{A}$

$\mathrm{A}=\frac{6}{4\times 10^3}=1.5\times 10^{-3}{\mathrm{m}}^2$

Converting this area from square meters to square centimeters:

$\mathrm{A}=1.5\times 10^{-3}{\mathrm{m}}^2=15{\mathrm{cm}}^2$