Question

Inside a conducting spherical shell of inner radius $3R$ and outer radius $5R$ , a point charge $Q$ is placed at a distance $R$ from the centre of the shell. The electric potential at the centre of the shell will be.

• A. $\frac{1}{4\pi {ϵ}_0}\frac{Q}{R}$
  
• B. $\frac{1}{4\pi {ϵ}_0}\frac{5Q}{6R}$
• C. $\frac{1}{4\pi {ϵ}_0}\frac{13Q}{15R}$
• D. $\frac{1}{4\pi {ϵ}_0}\frac{7Q}{9R}$

Answer 1

The correct option is C $\frac{1}{4\pi {ϵ}_0}\frac{13Q}{15R}$

From conservation of charge, the charge distribution on each surface is as shown in the figure.

Here in the figure , $1$ represents the point charge, $2$ represents inner surface of shell and $3$ represents the surface of outer shell.

Electric potential at centre due to point charge,
$V_1=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R}$

Electric potential at centre due to inner surface of shell,
$V_2=\frac{1}{4\pi {ϵ}_0}\frac{(-Q)}{3R}$

And, electric potential at centre due to outer surface of shell,
$V_2=\frac{1}{4\pi {ϵ}_0}\frac{\left(Q\right)}{5R}$

Net potential at the centre
$V=V_1+V_2+V_3$

$\Rightarrow V=\frac{1}{4\pi {ϵ}_0}[\frac{Q}{R}-\frac{Q}{3R}+\frac{Q}{5R}]$

$\Rightarrow V=\frac{1}{4\pi {ϵ}_0}\left[\frac{Q}{R}\left(\frac{15-5+3}{15}\right)\right]$

$\therefore V=\frac{1}{4\pi {ϵ}_0}\frac{13Q}{15R}$

Hence, option (c) is the correct answer.