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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
  • A$\frac{ C _{1} C _{2}}{\left( C _{1}+ C _{2}\right)} V$
  • B$\frac{\left( C _{1}+ C _{2}\right)}{ C _{1} C _{2}} V$
  • C$\left( C _{1}+ C _{2}\right) V$
  • D$\left( C _{1}- C _{2}\right) V$
JEE MAIN 2022, Medium
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