Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Additional Exercise,Chapter-2 Exercise 2.27, NCERT Solutions for Class 12 Physics
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?
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Capacitance of a charged capacitor, C₁ = 4μF = 4 x 10⁻6F
Supply voltage, V₁ = 200 V
Electrostatic energy stored in C₁ is given by,
E₁ = 1/2 C₁V₁2
= 1/2 x 4 x 10⁻6 x (200)2
= 8×10⁻2 J
Capacitance of an uncharged capacitor, C2 = 2μF = 2 x 10⁻6F
When C2 is connected to the circuit, the potential acquired by it is V2.
According to the conservation of charge, initial charge on capacitor C₁ is equal to the final charge on capacitors, C₁ and C2.
Therefore V2 (C₁+ C2) = C₁V₁
V2 x (4+2) x 10⁻6= 4 x 10⁻6 x 200
V2 = (400/3) V
Electrostatic energy for the combination of two capacitors is given by,
E2 = 1/2 (C₁+ C2 ) = C2 V22
= (2+4) x 10⁻6 x (400/3)2
= 5.33×10⁻2 J
Hence, amount of electrostatic energy lost by capacitor C₁ = E₁ — E2 = 0.08 – 0.0533 = 0.0267 = 2.67 x 10⁻2 J