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Question

A solid sphere of radius a has volume charge density p given by p=p1,0ra2 =p2,a2ra and electric potential at center is twice as electric potential at surface:

A
p1=2p2
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B
p1=p2
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C
p1=5p2
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D
p1=3p2
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Solution

The correct option is C p1=5p2
0ra2:ρ=ρ1
a2ra:ρ=ρ2
The system is equivalent to two solid spheres (i) radius a, density ρ2 (ii) radius a2, density (ρ1ρ2).
potential at O:Vo=14πεo.32.4πa33ρ2a+14πεo.32.4π3a38×(ρ1ρ2)×2a
=a2ρ22εo+a28εo(ρ1ρ2)=a28εo(ρ1+3ρ2)
Potential on the surface at P :
VP=14πεo.4πa33ρ2a+14πεo.4π3a38(ρ1ρ2)a
=a2ρ23εo+a2(ρ1ρ2)24εo=a224εo(8ρ2+ρ1ρ2)
a224εo(ρ1+7ρ2)
Vo=2Vp, then, 2(ρ1+7ρ2)=(3ρ1+9ρ2)
i.e., 2ρ1+14ρ2=3ρ1+9ρ2
ρ=5ρ2

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