Question

Figure shows a system of three concentric metal shells $\text{A, B}$ and $\text{C}$ with radii $a,2a$ and $3a$ respectively. Shell $\text{B}$ is earthed and shell$\text{C}$ is given a charge $Q$. Now, if shell $\text{C}$ is connected to shell $\text{A}$, then the final charge on the shell $\text{B}$ is equal to

• A. $-\frac{4Q}{13}$
• B. $-\frac{8Q}{11}$
• C. $-\frac{5Q}{3}$
• D. $-\frac{3Q}{7}$

Answer 1

The correct option is B $-\frac{8Q}{11}$

Let $q_1$ and $q_2$ be the charges on spheres $\text{A}$ and $\text{B}$ respectively.


When the shell $\text{A}$ and $\text{C}$ are connected by wire, their potentials become equal,
$V_A=V_C$

Substituting the values of $V_A$ and $V_C$,

$\Rightarrow \frac{K{q}_1}{a}+\frac{K{q}_2}{2a}+\frac{K(Q-q_1)}{3a}=\frac{K(q_1+q_2+Q-q_1)}{3a}$

$\Rightarrow 4q_1+q_2=0.....\left(1\right)$

When the shell $\text{B}$ is grounded,
$V_B=0$

$\Rightarrow \frac{K{q}_1}{2a}+\frac{K{q}_2}{2a}+\frac{K(Q-q_1)}{3a}=0$

$\Rightarrow q_1+3q_2=-2Q.....\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$ we get,

$q_2=-\frac{8Q}{11}$

Hence, option (b) is the correct answer.