A `4muF` capacitor is charged by a 200 V supply. It is then disconnected from the supply . And is connected to another uncharged `2muF` capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

Explanation

Here `C_(1) = 4 mu F = 4 xx 10^(-6) F, V_(1) = 200` volt.
Initial elctrostatic energy stored in `C_(1)` is `E_(1)`
`= (1)/(2) C_(1) V_(1)^(2) = (1)/(2) xx 4 xx 10^(6) xx 200 xx 200`
`E_(1) = 8 xx 10^(2)` Joule
When `4 mu F` capacitor is connected to uncharged capacitor of `2 mu F` charge flows and both acquire a common potential
`V = ("total charge")/("total capacity")`
`V = (C_(1) V_(1))/(C_(1) + C_(2)) = (4 xx 10^(-6) xx 200)/((4 + 2)10^(-6)) = (880)/(6)` volt
`:.` final electrostatic energy of both capacitors
`E_(2) = (1)/(2) (C_(1) + C_(2)) V^(2)`
`= (1)/(2) xx 6 xx 10^(-6) xx (800)/(6) xx (800)/(6)`
`E_(2) = 5.33 xx 10^(23)` joule
Energy dissipated in the form of heat adn electro magnetic radiation.
`E_(1) - E_(2) = 8 xx 10^(-2) - 5.33 xx 10^(-2)`
`= 2.67 xx 10^(-2)` joule.