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Question

# The charge per unit volume inside an insulating sphere of radius R is given as a function of distance from centre ρ(r)=cr , Where 0<r<R. The magnitude of electric field at a distance ′b′ from the centre of sphere will be

A
cπb4
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B
cb24ϵ0
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C
cbπϵ0
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D
cb4ϵ0
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Solution

## The correct option is B cb24ϵ0 Let us consider a shell of thickness dr at a distance ′r′ from centre. Volume of the shell, dv=(4πr2)dr (dv= surface area × thickness) The total charge enclosed in the shell will be dqin=ρdv ⇒dqin=crdv (∵ρ=cr) Let us construct a Gaussian sphere at a distance b from centre, thus net charge enclosed by this sphere: qin=∫dqin=∫crdv ⇒qin=∫b0cr(4πr2)dr or, qin=4πc∫b0r3dr or, qin=4πc[r44]b0 ∴qin=πcb4 Applying Gauss's law at spherical surface. ∮→E.d→A=qinϵ0 or EA=πcb4ϵ0 ⇒E(4πb2)=πcb4ϵ0 (∵A=4πb2) ⇒E=cb24ϵ0 Why this question?It intends to test your understanding in finding qinfor a nonuniformly charged sphere and applying Gauss lawTip: Electric field can be found by considering a gaussian surface (sphere), if we can obtain net charge enclosed by it.

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