Question

A sphere of 4cm radius is suspended with in a hollow sphere of 6 cm

Answer 1

Explanation:

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