JEE MAIN - Physics Hindi (2024 - 4th April Evening Shift - No. 15)
A charge $$q$$ is placed at the center of one of the surface of a cube. The flux linked with the cube is:
$$\frac{q}{2 \epsilon_0}$$
Zero
$$\frac{q}{4 \epsilon_0}$$
$$\frac{q}{8 \epsilon_0}$$
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