Question

A metal sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere Is charged to a potential of 3 esu with outer sphere connected to the earth, the charge on the inner sphere is

• A. 54 esu
• B. 1/4 esu
• C. 30 esu
• D. 36 esu

Answer 1

The correct option is D 36 esu

Let the charge of inner sphere$=+Q$

Then $-Q$ is introduced in outer sphere now, potential and inner sphere$=Q_1/R_1+Q_2/R_2$

Here $Q_1=$ charge on inner sphere

$R_1=$ Radius of inner sphere

$Q_2=$ Charge on outer sphere

$R_2=$ Radius of outer sphere

Given potential on inner sphere$=3esu$ of pot

$3esu=\frac{Q}{4}-\frac{Q}{6}\frac{Q}{12}=3Q=36esu$