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Question

A circular disc of radius 𝑅 carries surface charge density σ(r)=σo(1rR),where σo is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ϕ0 . Electric flux through another spherical surface of radiusR4 and concentric with the disc is ϕ.Then the ratio ϕ0ϕ is

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Solution


Let us choose an elementary ring of radius r and thickness dr as shown in the above figure .

So , the area of elementary ring dA = 2πrdr

ϕ0=dqε0=R0σ0(1R)2πrdrε0
ϕ=dqε0=R40σ0(1R)2πrdrε0
ϕ0ϕ=σ02πR0(1r2R)drσ02πR40(1r2R)dr
ϕ0ϕ=R22R23R232R23×64=325=6.40

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