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

Hint:

By Gauss law the flux depends only on enclosed charge.

Answer 1

The Correct Option is B

Step 1: Check with numbers:
Field at radius $r$ is $E=\dfrac{q}{4\pi\varepsilon_0r^2}$, and area $=4\pi r^2$.

Step 2: Multiply:
$\Phi=EA=\dfrac{q}{\varepsilon_0}$, independent of $r$.

Step 3: Pick:
Remains unchanged, option B.

Final Answer:
Flux depends only on the enclosed charge, which stays the same. \[ \boxed{\text{(B) }\text{remains unchanged}} \]