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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
Questions With Solutions

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Question : 27
Total: 37
A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Solution:  
C1 = 4 µF , V1 = 200 V , C2 = 2 µF , V2 = 0
So, common potential difference across the two capacitors after connection is
V =
C1V1+C2V2
C1+C2
 =
4×10−6×200+0
(4+2)×10−6
 = 133.33 V
Initially, total energy stored in capacitors before connection is
Ui =
1
2
C1V12 =
1
2
 × 4 × 10−6×2002 = 0.08 J
and total energy stored in capacitors after connection is
Uf =
1
2
(C1+C2)V2 =
1
2
 (4 + 2) × 10−6×133.332 or Uf = 0.053 J
So, energy lost due to connection is
ΔU = Uf−Ui = 0.053 - 0.08 or ΔU = - 0.027 J
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