NCERT Class XII Chapter
Electrostatic Potential and Capacitance
Questions With Solutions

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Question : 11
Total: 37
A 600 pF capacitor is charged by a 200 V 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?
Solution:  
C1 = 600 pF, V1 = 200 V , C2 = 600 pF , V2 = 0
On connecting charged capacitor to uncharged capacitor, the common potential V across the capacitors is
V =
C1V1+C2V2
C1+C2
=
600×1012×200+0
(600+600)×1012
or V = 100 V
Energy stored in capacitors before connection is
Ui =
1
2
C1
V12
+ 0 =
1
2
× 600 × 1012×2002 or Ui = 12 µJ
and energy stored in capacitors after connection is
Uf =
1
2
(C1+C2)
V2
=
1
2
(600 + 600) × 1012×1002 or Uf = 6 µJ
Hence the energy lost in the process is
Δ = UfUi = (6 - 12) µJ or ΔU = - 6 µJ
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