Question

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?

Answer 1

Capacitance of a charged capacitor,

Supply voltage, V₁ = 200 V 

Electrostatic energy stored in C₁ is given by,

Capacitance of an uncharged capacitor,

 When C₂ is connected to the circuit, the potential acquired by it is V_2. According to the conservation of charge, initial charge on capacitor C₁ is equal to the final charge on capacitors, C₁ and C_2.

Electrostatic energy for the combination of two capacitors is given by,

Hence, amount of electrostatic energy lost by capacitor C₁

= E₁ − E₂ = 0.08 − 0.0533 = 0.0267 = 2.67 × 10⁻² J