A charge q is enclosed by an imaginary Gaussian surface.If radius of surface is increasing at the rate drdt = K, then
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a
flux linked with surface is increasing at the rate dϕdt=K
b
flux linked with surface is decreasing at the rate dϕdt=-K
c
flux linked with surface is decreasing at the rate dϕdt=1K
d
flux linked with surface is qεo
answer is D.
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Detailed Solution
From Gauss' law ϕ=qenclosedεo in which qenclosed is the net charge inside an imaginary closed surface (a Gaussian surface) flux does not depend on the radius of imaginary enclosed surface.