A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 μF capacitor. The energy lost in the process is

The initial energy E1 is given byE1=12C1V12=12×4×10−6(200)2     =8×10−2JThe common potential V is given byV=q1+q2C1+C2=4×200+04+2=4003VThe final energy E2 is given byE2=12C1+C2V2    =126×10−6(400/3)2=5⋅33×10−2JLoss of energy =E1−E2                         =8×10−2−5⋅33×10−2=2⋅67×10−2J