A capacitor $C _{1}$ of capacitance $5\,\mu F$ is charged to a potential of $30\,V$ using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor $C _{2}$ of capacitance $10\,\mu F$ as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor $C _{2}$ is________ $\mu C$
JEE MAIN 2022, Diffcult
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Before closing the switch
$Q = C _{1} V _{0}=5 \times 30=150\,\mu C$
After closing the switch
$V =\frac{ Q }{ C _{1}+ C _{2}}=\frac{150}{10+5}=10\,V$
$Q _{2}= C _{2} V =10 \times 10=100\,\mu C$
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