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Question

A positively charged sphere of radius r0 carries a volume charge density ρ as shown in figure. A spherical cavity of radius r02 is then scooped out and left empty. C1 is the centre of the sphere and C2 is the centre of the cavity. What is the direction and magnitude of the electric field at point B?


A
ρr06ϵ0, towards left
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B
ρr06ϵ0, towards right
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C
17ρr054ϵ0, towards right
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D
17ρr054ϵ0, towards left
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Solution

The correct option is D 17ρr054ϵ0, towards left
Given,
Charge density or charge per unit volume=ρ
Radius of sphere =r0

If, Q is the total chrage on the sphere of radius r0, then
ρ=Q43πr30

Electric field on surface of the solid sphere at point B is given by

Ewhole sphere=Q4πϵ0r20=ρr03ϵ0...(1)

Electric field at a point outside the sphere at distance r is given by

Eo=Q4πϵ0r2=ρr303ϵ0r2

We can assume the spherical cavity to be made of material having charge density ρ.
So, using above formula, the electric field at point B due to spherical cavity can be calculated by taking, rc=r02 and r=32r0,

Ecavity=ρ(r02)33ϵ0(3r02)2...(2)

So the net electric field at point B,

EB=Ewhole sphereEcavity

Substituting the values from equation (1) and (2),
EB=ρr03ϵ0(towards left)ρ(r02)33ϵ0(3r02)2(towards right)

EB=17ρr054ϵ0, towards left.

Hence, option (a) is the correct answer.

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