A battery of `20V` is connected to a capacitor of capacitance `10 muF`. Now battery is removed and this capacitor is connected to a second uncharged capacitor. If the charge distributes equally on these two capacitors, the total energy stored in the two capacitors will be

Options

A. `250muJ`
B. `500muJ`
C. `750muJ`
D. `1000muJ`

Answer

D

Explanation

Since final charges are same, potential is same i.e.,
`C_(1)=C_(2)=10muF`
`V=(10xx20+10xx0)/(10+10)=10V`
`U_(f)=(1)/(2)(C_(1)+C_(2))V^(2)=(1)/(2)(10+10)(10)^(2)xx10^(-6)`
`=1000muJ`