Q52 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
Two point charges $2 q$ and $q$ are placed at vertex $A$ and centre of face $C D E F$ of the cube as shown in figure. The electric flux passing through the cube is :
- A$\frac{3 q}{2 \epsilon_{0}}$
- Correct answerB$\frac{3 q}{4 \epsilon_{0}}$
- C$\frac{q}{\epsilon_{0}}$
- D$\frac{3 q}{\epsilon_{0}}$
Solution
Using Gauss's law:
$\phi = \frac{Q_{in}}{\epsilon_0}$
Charge $2q$ is at vertex A of the cube. A vertex is shared by 8 cubes, so the flux contribution from $2q$ through this cube is $\frac{2q}{8\epsilon_0} = \frac{q}{4\epsilon_0}$.
Charge $q$ is at the centre of face CDEF. A face center is shared by 2 cubes, so the flux contribution from $q$ through this cube is $\frac{q}{2\epsilon_0}$.
Total flux:
$\phi = \frac{q/4 + q/2}{\epsilon_0} = \frac{3q}{4\epsilon_0}$
$\phi = \frac{Q_{in}}{\epsilon_0}$
Charge $2q$ is at vertex A of the cube. A vertex is shared by 8 cubes, so the flux contribution from $2q$ through this cube is $\frac{2q}{8\epsilon_0} = \frac{q}{4\epsilon_0}$.
Charge $q$ is at the centre of face CDEF. A face center is shared by 2 cubes, so the flux contribution from $q$ through this cube is $\frac{q}{2\epsilon_0}$.
Total flux:
$\phi = \frac{q/4 + q/2}{\epsilon_0} = \frac{3q}{4\epsilon_0}$
Hence, the correct answer is option: 'B'.