Question

A 4 μF capacitor is charged by a 200 V supply. It is then disconnectedfrom the supply, and is connected to another uncharged 2 μFcapacitor. How much electrostatic energy of the first capacitor islost in the form of heat and electromagnetic radiation?

Answer 1

It is given that the capacitance of a charged capacitor, C 1 =4μF, the capacitance of a uncharged capacitor, C 2 =2μF and the supply voltage, V 1 =200V.

The formula of electrostatic energy is,

E 1 = 1 2 C 1 V 1 2

Substitute the values.

E 1 = 1 2 ×4× 10 −6 × ( 200 ) 2 =8× 10 −2 J (1)

According to the law of conservation of charges,

V 2 ( C 1 + C 2 )= C 1 V 1

Substitute the values.

V 2 ×( 4+2 )× 10 −6 =4× 10 −6 ×200 V 2 = 400 3 V

The electrostatic energy for the combination of two capacitors is,

E 2 = 1 2 ( C 1 + C 2 ) V 2 2

Substitute the values.

E 2 = 1 2 ×( 2+4 )× 10 −6 × ( 400 3 ) 2 =5⋅33× 10 −2 J (2)

The amount of electrostatic energy lost by capacitor is,

E 1 − E 2 =8× 10 −2 −5⋅33× 10 −2 =2⋅67× 10 −2 J

Thus, the electrostatic energy of the first capacitor lost in the form of heat and energy is 2⋅67× 10 −2 J.