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 mu F` capacitor. How much electrostatic energy of the first capacitor is disspated in the form of heat and electromagnetic radiation ?

Explanation

When an uncharged capacitor is connected across a charged capacitor, the charged capacitor discharges through the uncharged capacitor, till both the capacitors attain equal potential difference between their plates.
Electrostatic energy stored in the capacitor initially:
`U_(i)=(1)/(2)CV^(2)=(1)/(2)(4xx10^(-6))(200)^(2)=8xx10^(-2)J`
Common potential after the charged capacitor is connected with uncharged capacitor:
`V=(C_(1)V_(1))/(C_(1)+C_(2))=(4xx10^(-6)xx200)/(4xx10^(-6)+2xx10^(-6))=(800)/(6)V`
Final electrostatic energy stored in both the capacitors
`U_(f)=(1)/(2)(C_(1)+C_(2))V^(2)=(1)/(2)(6xx10^(-6))((800)/(6))^(2)=5.33xx10^(-2)J`
While sharing charge, the energy dissipated in form of heat and electromagnetic radiation can be calculated as
`DeltaU=U_(i)-U_(f)=(8-5.33)xx10^(-2)=2.62xx10^(-2)J`