A body of capacity 4 μ F is charged to 80 V and another body of capacity 6 μ F is charged to 30V. When they are connected the energy lost by 4 μ F capacitor is

Initial energy of body of capacitance 4 μ F isUi=12×(4×10−6) (80)2=0.0128JFinal potential on this body after connection is V=4×80+6×304+6=50V. So final energy on it Uf=12×4×10−6(50)2=0.005 JEnergy lost by this body =Ui−Uf=7.8mL