A $10\, \mu F$ capacitor is fully charged to a potential difference of $50\, V$. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes $20\, V$. The capacitance of the second capacitor is......$\mu F$
JEE MAIN 2020, Diffcult
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Initially
Charge on capacitor 10 $\mu F$
$Q=C V=(10 \mu F)(50 V)$
$Q =500 \mu C$
Final Charge on $10 \mu F$ capacitor $Q = CV =(10 \mu F )(20 V )$
$Q =200 \mu C$
From charge conservation,
Charge on unknown capacitor
$C =500 \mu C -200 \mu C =300 \mu C$
$\Rightarrow$ Capacitance $( C )=\frac{ Q }{ V }=\frac{300 \mu C }{20 V }=15 \mu F$
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