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

Answer

15

Explanation

`q _(1) = 10 xx 50 = 500 mu C , C _(1) = 10 mu F,`
` C _(2) = ? , q _(2) = 0`
As, `V = ( q _(1) + q _(2))/( C _(1) + C_(2))`
` C _(1) + C _(2) + (q _(1) + q _(2))/( V) = ( 500 + 0)/( 20) = 25 mu F`
`C _(2) = 25 - C _(1) = 25 - 10 = 15 mu F`