Question

# A circular disc of radius 𝑅 carries surface charge density σ(r)=σo(1−rR),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=σ02π∫R0(1−r2R)drσ02π∫R40(1−r2R)dr ϕ0ϕ=R22−R23R232−R23×64=325=6.40

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