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Question

sphere of 4 cm radius is suspended within a hollow sphere of radius 6cm the inner sphere is charged to a potential 3S you when the outer sphere is earthed the charge on the inner sphere is

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Solution

Let charge on inner sphere is +Q
then, -Q is induced in the surface of outer sphere.
now, potential on inner sphere = Q₁/r₁ + Q₂/r₂
Here, Q₁ is the charge on inner sphere
r₁ is the radius of inner sphere
Q₂ is the charge on outer sphere
r₂ is the radius of outer sphere.

Given, potential on inner sphere = 3 esu of potential
r₁ = 4cm
r₂ = 6cm
∴ 3 esu = Q/4 - Q/6
⇒ Q[1/4 - 1/6 ] = 3
⇒Q/12 = 3
⇒ Q = 36 esu

Hence, charge = 36 esu of charge

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