A \(600\,pF\) capacitor is charged by \(200\,V\) supply. It is then disconnected from the supply and is connected to another uncharged \(600\,pF\) capacitor. Electrostatic energy lost in the process is \(.........\,\mu J\).

\(\text { Initial energy }=\frac{1}{2} CV ^2\)

\(=\frac{1}{2} \times 600 \times 10^{-12} \times(200)^2=12\,\mu J\)

When connected to another uncharged capacitor

Charge will be equally distributed on identical capacitor

\(Q ^{\prime}=\frac{ Q }{2}=6 \times 10^{-8}\)

Final energy \(=2 \times \frac{ Q ^{\prime 2}}{2 C }=\frac{ Q ^{\prime 2}}{ C }\)

\(\frac{\left(6 \times 10^{-8}\right)^2}{600 \times 10^{-12}}=6\,\mu J\)

Energy lost = Initial energy - Final energy

\(=(12-6)\,\mu\, J =6\, \mu\, J\)