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Question

A ball of radius R carries a positive charge whose volume charge density depends only on the distance r from the ball's centre as: ρ=ρ0(1−rR) where ρ0 is constant. Assume ϵ as the permittivity of the ball.Then the magnitude of the electric field as a function of the distance r outside the ball is given by :

A
E=ρ0R38ϵr2
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B
E=ρ0R312ϵr2
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C
E=ρ0R316ϵr2
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D
E=ρ0R324ϵr2
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Solution

The correct option is B E=ρ0R312ϵr2Using Gauss's law,∫→E.^nds=QenclosedϵFor field at r outside the sphere, Qenclosed=∫R0ρ4πr2dr=4πρ0∫R0(1−rR)r2dr=4πρ0[R33−R44R]=4πρ0R312=πρ0R33thus, E.4πr2=πρ0R33ϵE=ρ0R312ϵr2

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