JEE Physics 2025 Question with Solution

Given: A line charge of length $\frac{a}{2}$ is kept at the center of edge $BC$ of a cube of edge length $a$. The linear charge density is $\lambda$.

Find: The total electric flux through all faces of the cube.

Use Gauss's law:

$$\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$

The line charge lies on an edge of the cube. An edge is common to four adjacent cubes, so the given cube encloses only one-fourth of the total charge on the line segment.

Total charge on the line segment:

$$q = \lambda \times \frac{a}{2} = \frac{\lambda a}{2}$$

Charge effectively enclosed by the cube:

$$q_{\text{enclosed}} = \frac{1}{4} \times \frac{\lambda a}{2} = \frac{\lambda a}{8}$$

Therefore, the total electric flux through all faces of the cube is:

$$\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda a}{8\varepsilon_0}$$

Therefore, the correct option is A.

Given: The charged segment is placed at the center of an edge of the cube.

Find: Why the enclosed charge is not the full line charge.

Imagine placing four identical cubes around that edge so that the charged segment becomes an interior line of the combined arrangement. By symmetry, the flux contribution is shared equally among the four cubes.

The full line charge is:

$$q = \frac{\lambda a}{2}$$

Hence each cube accounts for:

$$\frac{q}{4} = \frac{1}{4}\left(\frac{\lambda a}{2}\right) = \frac{\lambda a}{8}$$

Applying Gauss's law to the given cube:

$$\Phi = \frac{1}{\varepsilon_0} \cdot \frac{\lambda a}{8} = \frac{\lambda a}{8\varepsilon_0}$$

This also shows why the intermediate result $\frac{\lambda a}{2\varepsilon_0}$ corresponds to the flux associated with the entire line charge, not the share enclosed by the single cube.