A 600 pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected

to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process. ?

Capacitance of a charged capacitor, C_{1}= 600 x 10^{-12 }F

Supply voltage, *V*_{1} = 200 V

Electrostatic energy stored in *C*_{1 }is given by,

Capacitance of an uncharged capacitor, C_{2}= 600pF

When *C*_{2} is connected to the circuit, the potential acquired by it is *V*_{2}.

According to the conservation of charge, initial charge on capacitor *C*_{1} is equal to the final charge on capacitors, *C*_{1} and *C*_{2}.

Electrostatic energy for the combination of two capacitors is given by,

= 6 x 10^{-6} J

Hence, amount of electrostatic energy lost by capacitor *C*_{1}

= *E*_{1} − *E*_{2}

= 6 x 10^{-6}J

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