Question

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` afrte reconnection?

Answer 1

Here, `C_(1) = C_(2) = 600 pF = 600xx10^(-12)F = 6xx10^(-10) F, V_(1) = 200V, V_(2) = 0`
Loss of energy `= (C_(1) C_(2) (V_(1) - V_(2))^(2))/(2(C_(1) + C_(2))) = (6xx10^(-10)xx6xx10^(-10)(200-0)^(2))/(2(6xx10^(-10)+6xx10^(-10))) = (36xx10^(-20)xx4xx10^(4))/(24xx10^(-10)) = 6xx10^(-6)J`