Q.

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.
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