Question

If the radius of the spherical Gaussian surface is increased then the electric flux due to a point charge enclosed by the surface

• A. decreases
• B. remains unchanged
• C. increases
• D. is zero

Hint:

Electric flux through a closed surface depends only on enclosed charge, not on surface area or radius.

Answer 1

The Correct Option is B

Step 1: Gauss’s law.
According to Gauss’s law, electric flux through a closed surface is \[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}. \]
Step 2: Dependence on radius.
The flux depends only on the charge enclosed by the surface and not on the size or shape of the Gaussian surface.
Step 3: Conclusion.
Since the enclosed charge remains the same, the electric flux remains unchanged even if the radius increases.