Question

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

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

$q_1$ and $q_2$ are two unknowns

$V_A=V_C\therefore \frac{K{q}_1}{a}+\frac{K{q}_2}{2a}+\frac{K(Q-q_1)}{3a}=\frac{K(q_1+q_2+Q-q_1)}{3a}or4q_1+q_2=0\dots \dots \left(1\right)V_B=0\therefore \frac{K{q}_1}{2a}+\frac{K{q}_2}{2a}+\frac{K(Q-q_1)}{3a}=0$
or $q_1+3q_2=-2Q\dots \dots \left(2\right)$
Solving these two equations we get,
$q_2=-\frac{8Q}{11}$