A `10 muF` capacitor is charged to a potential difference of `50 V` and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes `20` volt. The capacitance of second capacitor is

Options

A. `10muF`
B. `20muF`
C. `30muF`
D. `15muF`

Answer

D

Explanation

Charge of first capacitor, `q=CV = 10xx50=500muC`
Common potentail `=("Total charge")/("Total capacitance")`
`20=(500)/(C_(1)+C_(2)) rArr 20 =(500)/(10+C_(2))`
`C_(2)=15muF`