A `600 pF` capacitor is charged by a `200 V` supply. It is then disconnected from the supply and is connected to another uncharged `600 pF` capacitor. What is the common potential in `V` and energy lost in `J` after reconnection?

Options

A. `100, 6xx10^(-6)`
B. `200, 6xx10^(-5)`
C. `200, 5xx10^(-6)`
D. `100, 6xx10^(-5)`

Answer

A

Explanation

`V_(C)=(C_(1)V_(1))/(C_(1)+C_(2))=100 V`
Energy lost `=U_(i)-U_(f)`
`=1/2C_(1)V_(1)^(2)-1/2(C_(1)+C_(2))V_(c)^(2)`
`=6xx10^(-6)J`.