A charge $q$ is placed at the center of one of the surface of a cube. The flux linked with the cube is:

1. $\frac{q}{2 \epsilon_0}$
2. $\frac{q}{8 \epsilon_0}$
3. Zero
4. $\frac{q}{4 \epsilon_0}$

Solution

From
$\begin{aligned} & 2 \phi=\frac{\mathrm{q}}{\epsilon_0} \\ & \phi=\frac{\mathrm{q}}{2 \epsilon_0}\end{aligned}$

Asked in: JEE Main 2024 (04 Apr Shift 2)

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