Question

In the given figure, radius of conducting shell 'A', 'B' and 'C' are $R$, 2$R$ and 3$R$ respectively. Charge on conducting shell 'A', 'B' and 'C' are 3$Q$, -2$Q$ and $Q$ respectively. Charge flown from 'A' to 'C', when switch 'S' is closed is

• A. $\frac{3Q}{2}$
• B. $\frac{-Q}{2}$
• C. $\frac{5Q}{2}$
• D. $\frac{-3Q}{2}$

Answer 1

The correct option is C $\frac{5Q}{2}$

Let $q$ be the charge that flows from A to C.
So charge on A changes to $3Q-q$ and charge on B changes to $Q+q$.

$V_A=V_C\frac{K(3Q-q)}{R}+\frac{-2KQ}{2R}+\frac{K(Q+q)}{3R}=\frac{K(3Q-q)}{3R}+\frac{-2KQ}{3R}+\frac{K(Q+q)}{3R}q=\frac{5Q}{2}$