Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Chapter-2 Exercise 2.11, NCERT Solutions for Class 12 Physics
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?
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Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
E₁=1/2 x CV2
= 1/2 x (600 x 10⁻12) x (200)2 J = 1.2 x 10⁻5 J
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (Ceq) of the combination is given by,
1/Ceq=1/C + 1/C
=> 1/Ceq =1/600 + 1/600=2/600 =1/300
=> Ceq =300 pF
New electrostatic energy can be calculated as
E2=1/2 Ceq V2
=1/2 x 300 x (200)2 J = 0.6 x 10⁻5 J
Loss in electrostatic energy = E₁ – E2
= 1.2 x 10⁻5 – 0.6 x 10⁻5 J = 0.6 x 10⁻5J = 6x 10⁻6J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J .