Question

Three concentric spherical shells have radii r, 2r, and 3r. Innermost and outermost shells are earthed. Now charges on the shells are $q_1$, $q_2$ and $q_3$ respectively. Then :

• A. $q_1+q_3=-q_2$
• B. $q_1=-\frac{q_2}{4}$
• C. $\frac{q_3}{q_1}=3$
• D. $\frac{q_3}{q_2}=-\frac{1}{3}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $q_1+q_3=-q_2$
B $q_1=-\frac{q_2}{4}$
C $\frac{q_3}{q_1}=3$

$\text{Step 1: General Formula of Potential of Sphere of Radius R}$

At outside point at a distance $x$ from its centre,

$V_x=\frac{Kq}{x}$

At inside point, it is constant, same as that over its surface,

$V_{inside}=\frac{Kq}{R}$

$\text{Step 2: Potential at sphere (1) and sphere (3)}$ $\text{[Ref. Fig.]}$

Potential being a scalar quantity, it will be added due to all the spheres at a point.

So, Potential at sphere $\left(1\right)$

$V_1=V_{Self}+V_{BySph.2}+V_{BySph.3}$

$\Rightarrow V_1=\frac{K{q}_1}{r}+\frac{K{q}_2}{2r}+\frac{K{q}_3}{3r}$ $....\left(1\right)$ (For spheres $2$ & $3$, sphere $1$ is a inside point, so using formula accordingly)

Potential at sphere $\left(3\right)$

$V_3=V_{BySph.1}+V_{BySph.2}+V_{Self}$

$\Rightarrow V_3=\frac{K{q}_1}{3r}+\frac{K{q}_2}{3r}+\frac{K{q}_3}{3r}$ $....\left(2\right)$ (For spheres $1$ & $2$, sphere $3$ is a outside point, so using formula accordingly)

$\text{Step 3: Earthing of sphere (1) and sphere (3)}$

On earthing potential becomes zero:

Therefore, from equation $\left(1\right)$, $V_1=0$

$\Rightarrow \frac{K{q}_1}{r}+\frac{K{q}_2}{2r}+\frac{K{q}_3}{3r}=0$

$\Rightarrow 6q_1+3q_2+2q_3=0$ $.....\left(3\right)$

Therefore, from equation $\left(2\right)$, $V_3=0$

$\Rightarrow \frac{K{q}_1}{3r}+\frac{K{q}_2}{3r}+\frac{K{q}_3}{3r}=0$

$\Rightarrow q_1+q_2+q_3=0$ $.....\left(4\right)$

$q_1+q_3=-q_2$

On solving (3) and (4), we get:

$q_1=-\frac{q_2}{4}$ and $\frac{q_3}{q_1}=3$

Hence, option $A,B,C$ are correct.