Question

A sphere of $4 \,cm$ radius is suspended within hollow sphere of $6 \,cm$ radius. The inner sphere is charged to a potential $3 \,esu$, when the outer sphere is earthed. The charge on the inner sphere is :

• A. 54 esu
• B. 27 esu
• C. 30 esu
• D. 36 esu

Answer 1

Answer: (D)

The sphere of radius $=4 \,cm =4 \times 10^{-2} m$
hollow sphere $=6 \,cm =6 \times 10^{-2} m$
inner sphere charged to a potential $3$ e.s.u.
lesu $=300\, V$
lesu $=3 \times 10^{-10} C$
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 $=3$ esu of potential
$r _{1}=4 \,cm$
$r _{2}=6\, cm$
$ 3$ esu $=\frac{ Q }{4}-\frac{ Q }{6}$
$\Rightarrow Q \left[\frac{1}{4}-\frac{1}{6}\right]=3$
or, $\frac{ Q }{12}=3$
or, $Q =36$ esu
Hence, charge $=36$ esu of charge.