A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another unchanged 600pF capacitor. How much electrostatic energy is lost in the process.

Explanation

Here `C_(1) 600 pE ,C_(2) = 600 pF, V_(1)=200 V , V_(2) = 0` V
From given formula
`:.` Loss in electrostatic energy
`DeltaU = (C_(1)C_(2)(V_(1)-V_(2))^2)/(2(C_(1)+C_(2)))`
`=(600xx10^(-12)xx600xx10^(-12)(200-0))/(2(600xx10^(12)600xx10^(-12)))`
`=(36xx10^(-20)xx200)/(2xx1200xx10^(-12))`
`= 6xx10^(-6)` J Loss in energy.