Question

A sphere of $4$ cm radius is suspended within a hollow sphere of $6$ cm radius. The inner sphere is charged to a potential $3$ e.s.u. When the outer sphere is earthed, the charge on the inner sphere is:

• A. $54$ e.s.u.
• B. $\frac{1}{4}$ e.s.u.
• C. $30$ e.s.u.
• D. $36$ e.s.u.

Answer 1

The correct option is D $36$ e.s.u.

The sphere of radius $=4cm$ $=4\times {10}^{-2}m$

hollow sphere $=6cm$ $=6\times {10}^{-2}m$

inner sphere charged to a potential $3$ e.s.u.

$1esu=300V$

$1esu=3\times {10}^{-10}C$

Let, Charge on inner sphere is $+Q$, then, $-Q$ is induced in the surface of outer sphere.

Now, potential on inner sphere $=\frac{Q_1}{r_1}+\frac{Q_2}{r_2}$

Here, $Q_1$ is the charge on inner sphere.

$r_1$ is the radius of inner sphere.

$Q_2$ is the charge on outer sphere.

$r_2$ $=$ the radius of the outer sphere.

Given, Potential on inner sphere $=3esu$ of potential

$r_1=4cm$

$r_2=6cm$

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

$\Rightarrow Q[\frac{1}{4}-\frac{1}{6}]=3$

or, $\frac{Q}{12}=3$

or, $Q=36esu$

Hence, charge $=36esu$ of charge.