Question

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

If radius of surface is increasing at the rate $\frac{d r}{d t}=K$, then

• A. flux linked with surface is increasing at the rate $\frac{d \phi}{d t}=K$
• B. flux linked with surface is decreasing at the rate $\frac{d \phi}{d t}=-K$
• C. flux linked with surface is decreasing at the rate $\frac{d \phi}{d t}=\frac{1}{K}$
• D. flux linked with surface is $\frac{q}{\varepsilon_{0}}$

Answer 1

Answer: (D)

From Gauss' law $\phi=\frac{q_{\text {enclosed }}}{\varepsilon_{0}}$ 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.