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

Explanation

Given `C_(1)=4muF=4xx10^(-6),V_(1)=200V`
Initial energy of first capacitor `U_(1)=(1)/(2)C_(1)V_(1)^(2)=(1)/(2)xx(4xx10^(-6))xx(200)^(2)=8xx10^(-2)J`
when another `C_(2)=2muF`, uncharged capacitor is connected across first capacitor,
Common potential, `V=(q_(1)+q_(2))/(C_(1)+C_(2))=(C_(1)V_(1)+0)/(C_(1)+C_(2))=(4xx10^(-6))/((4+2)xx10^(-6))=(400)/(3)`Volt
Final energy, `U_(2)=(1)/(2)(C_(1)+C_(2))V^(2)=(1)/(2)xx(4xx2)xx10^(-6)xx((400)/(3))^(2)=(16)/(3)xx10^(-2)J=5.33xx10^(-2)J`
Energy loss, `DeltaU=U_(1)-U_(2)=8xx10^(-2)-5.33xx10^(-2)=2.67xx10^(-2)J`.