Question

A 600pF 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?

Answer 1

Given: The capacitance of the capacitor is 600 pF and the supply voltage is 200 V.

The electrostatic energy stored in the capacitor is given as,

E= 1 2 C V 2

where, the electrostatic energy stored in the capacitor is E, capacitance of the capacitor is C, and the potential difference across the capacitor is V

By substituting the given values in the above equation, we get,

E= 1 2 ×600 pF×( 1 F 10 12  pF )× ( 200 ) 2 =1.2× 10 −5  J

If supply is disconnected from the capacitor and connected to another capacitor, then capacitors are in series.

Total capacitance of the combination of capacitors when connected in series is given as,

1 C t = 1 C 1 + 1 C 2

where, total capacitance is C t , C 1 and C 2 are the capacitance of each capacitor.

By substituting the given values in the above equation, we get,

1 C t = 1 600 + 1 600 1 C t = 2 600 C t =300 pF

Total capacitance of the combination when connected in series is 300 pF.

New electrostatic energy stored in the capacitor is given as,

E'= 1 2 C' V 2

where, the electrostatic energy stored in the capacitor is E', capacitance of the capacitor is C', and the potential difference across the capacitor is V

By substituting the given values in the above equation, we get,

E'= 1 2 ×300 pF×( 1 F 10 12  pF )× ( 200 ) 2 =0.6× 10 −5  J

Loss of electrostatic energy ( ΔE ) in the process is,

ΔE=E−E' =1.2× 10 −5 −0.6× 10 −5 =0.6× 10 −5  J =6× 10 −6  J

Therefore, the loss of electrostatic energy in the process is 6× 10 −6  J.