A charge \( q \) is enclosed by an imaginary Gaussian surface.
If radius of surface is increasin... VIDEO
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 tabe \( \frac{d \phi}{d t}=-K \)
(C) flux linked with surface is decreasing at the rate
\[
\frac{d \phi}{d t}=\frac{1}{K}
\]
(D) flux linked with surface is \( \frac{\varphi}{E_{0}} \)
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