JEE Main 2025 — Electrostatics Question with Solution

To find the net flux through a Gaussian cube when an infinitely long wire with a uniform linear charge density $\lambda =2\text{ nC/m}$ passes through the cube's diagonal corners, we use Gauss's Law. The relevant expression is:

$\varphi =\frac{q_{\text{enc}}}{{\epsilon }_0}$

Here, $q_{\text{enc}}$ is the charge enclosed by the Gaussian surface, and ${\epsilon }_0$ is the permittivity of free space.

Given:

The wire passes through two opposite corners of the cube.

The side length of the cube, $a=\sqrt{3}\text{ cm}=\sqrt{3}\times 10^{-2}\text{ m}$.

The length of the wire inside the cube is equal to its diagonal, which is $\sqrt{3}\times a$.

Thus, the charge enclosed $q_{\text{enc}}$ is:

$q_{\text{enc}}=\lambda \times \text{(length of wire inside the cube)}=\lambda \times \sqrt{3}\times a$

Substitute the given values:

$\lambda =2\times 10^{-9}\text{ C/m},a=\sqrt{3}\times 10^{-2}\text{ m}$

Calculating:

$q_{\text{enc}}=2\times 10^{-9}\times \sqrt{3}\times \sqrt{3}\times 10^{-2}$

Now apply Gauss's Law for net flux:

$\varphi =\frac{\lambda \cdot \sqrt{3}\cdot a}{{\epsilon }_0}$

Substitute ${\epsilon }_0=\frac{1}{4\pi \times 9\times 10^9}$:

$\varphi =2\times 10^{-9}\times \sqrt{3}\times \sqrt{3}\times 10^{-2}\times 36\pi \times 10^9$

$\varphi =2.16\pi {\text{ Nm}}^2{\text{C}}^{-1}$

Thus, the net flux through the Gaussian cube is $2.16\pi {\text{ Nm}}^2{\text{C}}^{-1}$.