Q142 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
If the radius of the spherical Gaussian surface is increased then the electric flux due to a point charge enclosed by the surface
- Aincreases
- Correct answerBremains unchanged
- Cis zero
- Ddecreases
Solution
According to Gauss's law, the total electric flux through a closed surface is given by $\Phi = \dfrac{q_{enclosed}}{\epsilon_0}$.
The electric flux depends only on the net charge enclosed by the surface and is independent of the size or shape of the Gaussian surface.
When the radius of the spherical Gaussian surface is increased, the charge enclosed by it remains the same.
Therefore, the electric flux remains unchanged.
Answer: remains unchanged
The electric flux depends only on the net charge enclosed by the surface and is independent of the size or shape of the Gaussian surface.
When the radius of the spherical Gaussian surface is increased, the charge enclosed by it remains the same.
Therefore, the electric flux remains unchanged.
Answer: remains unchanged
Hence, the correct answer is option: 'B'.