Question

A metal sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to potential $3$ esu and the outer sphere is earthed. 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

Given a metal sphere of radius 4 cm is suspended within a hollow sphere of radius 6 centimeter. we have to determine charge on linear surface potential of inner surface is zero.

let charge on inner sphear is + q.

then; - Q charge is induced in the surface of outer sphere.

Now potential on inner sphere.$=\frac{1}{4\pi {\epsilon }_0}(\frac{q_1}{r_1}+\frac{q_2}{r_2})$

hance,$Q_1=$charg on inner sphear

Given a metal sphere of radius 4 cm is suspended within a hollow sphere of radius 6 centimeter. we have to determine charge on linear surface potential of inner surface is zero.

let charge on inner sphere is + q.

then; - Q charge is induced in the surface of outer sphere.

Now potential on inner sphere.$=\frac{1}{4\pi {\epsilon }_0}(\frac{q_1}{r_1}+\frac{q_2}{r_2})$

hence,$Q_1=$charge on inner sphere

$r_1=$radius of inner sphere

$q_2\&r_2$ is charge & radios of outer shell respectively.

given potential on inner sphear =3esu

$r_1=$4cm

$r_2=$6cm

$\therefore 3=\frac{1}{4\pi {\epsilon }_0}[\frac{Q_1}{r_1}-\frac{Q}{r_2}]$

in C.G.S. systum

$3=\frac{Q}{4}-\frac{Q}{6}=\frac{2Q}{24}-\frac{Q}{12}Q=36esu.sotheoptionis\left(d\right).$