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

A

$$\frac{C_1 C_2}{(C_1+C_2)}V$$

B

$$\frac{(C_1+C_2)}{C_1 C_2}V$$

C

$$(C_1+C_2)V$$

D

$$(C_1-C_2)V$$