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, C1= 600 x 10-12 F

Supply voltage, V1 = 200 V

Electrostatic energy stored in C1 is given by,

Capacitance of an uncharged capacitor, C2= 600pF

When C2 is connected to the circuit, the potential acquired by it is V2.

According to the conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.

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

 = 6 x 10-6 J

Hence, amount of electrostatic energy lost by capacitor C1

= E1  E2

= 6 x 10-6J

  • 74
What are you looking for?