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
JEE MAIN 2022, Medium
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Charge on capacitor $C _{2}$
$=\frac{ C _{2} \times Q _{\text {total }}}{ C _{\text {total }}}=\frac{ C _{2}\left[ C _{1} V \right]}{ C _{1}+ C _{2}}=\frac{ C _{1} C _{2} V }{ C _{1}+ C _{2}}$
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