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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 :

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$\frac{(C_{1} + C_{2})}{C_{1} C_{2}} V$

$(C_{1} + C_{2}) V$

$\frac{C_{1} C_{2}}{(C_{1} + C_{2})} V$

$(C_{1} - C_{2}) V$

answer is A.

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Detailed Solution

Charge on capacitor $C_{2} = \frac{C_{2} \times Q_{t o t a l}}{Q_{t o t a l}} = \frac{C_{2} [C_{1} V]}{C_{1} + C_{2}} = \frac{C_{1} C_{2} V}{C_{1} + C_{2}}$

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