When the capacitor of 600 pF is charged by 200V supply, the electrostatic energy stored =1.2CV2 =
1
2
×600×10−12×(200)2 =
1
2
×600×10−12×200×200 =12×10−6J When it is connected to 600 pF uncharged capacitor, the electrostatic energy will be equally divided in the both the capacitors. So, the energy loss in first capacitor =