Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)
- A$600$
- B$200$
- C$400$
- D$100$
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