Two capacitors having capacitance $C_{1}$ and $C_{2}$ respectively are connected as shown in the 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
  $(C_{1}-C_{2}) V$