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?

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