Question

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

If radius of surface is increasing at a rate `(dr)/(dt)= K`, then
A. flux linked with surface is increasing at a rate `(dphi)/(dt)=K`
B. flux linked with surface is decreasing at a rate `(dphi)/(dt)=-K`
C. flux linked with surface is decreasing at a rate `(dphi)/(dt)=(1)/(K)`
D. flux linked with surface is `(q)/(epsilon_(0))`

Answer 1

Correct Answer - d
From the Gauss, law `phi=(q_(enclosed))/(epsilon_(0))`
in which `q_(enclosed)` is the net charge inside an imaginary closed surface (a Gsussian surface) flux does not depend on the radius of imaginary enclosed surface.