JEE Physics 2026 Question with Solution

Given: Two point charges $2q$ and $q$ are placed at a vertex and at the centre of a face of a cube.

Find: The total electric flux through the cube.

Using Gauss's law, the total electric flux through a closed surface is

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

When a charge lies on the boundary, only the symmetric fraction belonging to the cube is counted.

For the charge $2q$ at vertex A, the charge is shared by $8$ identical cubes. Hence its contribution is

$$\Phi_v = \frac{1}{8}\cdot \frac{2q}{\varepsilon_0} = \frac{q}{4\varepsilon_0}$$

For the charge $q$ at the centre of face CDEF, the charge is shared by $2$ identical cubes. Hence its contribution is

$$\Phi_f = \frac{1}{2}\cdot \frac{q}{\varepsilon_0} = \frac{q}{2\varepsilon_0}$$

Therefore, total flux through the cube is

$$\Phi = \frac{q}{4\varepsilon_0} + \frac{q}{2\varepsilon_0} = \frac{3q}{4\varepsilon_0}$$

Therefore, the electric flux passing through the cube is $\frac{3q}{4\varepsilon_0}$. The correct option is B.