The Total solution for NCERT class 6-12
A 4 µF capacitor is charged by a 200 Vsupply. It is then disconnected from the supply, and is connected to anotheruncharged 2 µF capacitor. How much electrostatic energy of the first capacitoris lost in the form of heat and electromagnetic radiation?
Capacitance of a chargedcapacitor,
Supply voltage, V1 =200 V
Electrostatic energy stored in C1 isgiven by,
Capacitance of an unchargedcapacitor,
When C2 isconnected to the circuit, the potential acquired by it is V2.
According to the conservation of charge,initial charge on capacitor C1 is equal to thefinal charge on capacitors, C1 and C2.
Electrostatic energy for the combination of twocapacitors is given by,
Hence, amount of electrostatic energy lostby capacitor C1
= E1 − E2
= 0.08 − 0.0533 = 0.0267
= 2.67 × 10−2 J