Question

A charge $q$ is enclosed by an imaginary Gaussian surface.

If radius of surface is increasing at the rate $\frac{dr}{dt}=K$, then

• A. flux linked with surface is increasing at the rate $\frac{d\varphi }{dt}=K$
• B. flux linked with surface is decreasing at the rate $\frac{d\varphi }{dt}=-K$
• C. flux linked with surface is decreasing at the rate $\frac{d\varphi }{dt}=\frac{1}{K}$
• D. flux linked with surface is $\frac{q}{{\epsilon }_0}$

Answer 1

Answer: (D)

From Gauss' law $\varphi =\frac{q_{\text{enclosed }}}{{\epsilon }_0}$ in which $q_{\text{enclosed }}$ is the net charge inside an imaginary closed surface (a Gaussian surface) flux does not depend on the radius of imaginary enclosed surface.