Q 12-02-113JEE MainJEE Main 2022 (26 Jun, Shift 1)Easy
Two capacitors having capacitance $C_1$ and $C_2$ respectively are connected as shown in figure. Initially, capacitor $C_1$ is charged to a potential difference $V$ volt by a battery. The battery is then removed and the charged capacitor $C_1$ is now connected to uncharged capacitor $C_2$ by closing the switch $S$. The amount of charge on the capacitor $C_2$, after equilibrium, is
Answer: (A) $\dfrac{C_1C_2}{C_1+C_2}V$
Charge $C_1V$ is shared until both capacitors have the same voltage:
$$V' = \frac{C_1V}{C_1+C_2},\qquad Q_2 = C_2V' = \frac{C_1C_2}{C_1+C_2}V$$
Solution by Sreeraj P, M.Sc Physics