A $2\, \mu F$ capacitor $C _{1}$ is first charged to a potential difference of $10\, V$ using a battery.Then the battery is removed and the capacitor is connected to an uncharged capacitor $C _{2}$ of $8\, \mu F$. The charge in $C _{2}$ on equilibrium condition is $\ldots\,\mu C$. (Round off to the Nearest Integer)
JEE MAIN 2021, Diffcult
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$20=\left( C _{1}+ C _{2}\right) V \Rightarrow V =2\, volt$
$Q _{2}= C _{2} V =16\, \mu C$
$=16$
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