Question

A 6 μF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer 1

Data: 

C = 6 µF = 6 × 10⁻⁶ F = C₁, V = 300 V, C₂ = 3 µF

The capacitor's electrostatic energy

= `1/2 CV^2` 

= `1/2(6 xx 10^-6)(300)^2`

= 3 × 10⁻⁶ × 9 × 10⁴

= 0.27 J

This capacitor has a charge on it.

Q = CV = (6 × 10⁻⁶)(300)

= 1.8 mC

When two capacitors of capacitances C₁ and C₂ are connected in parallel, the equivalent capacitance C

= C₁ + C₂

= 6 + 3

= 9 µF

= 9 × 10⁻⁶ F

By conservation of charge, Q = 1.8 C.

∴ The energy of the system = `Q^2/(2C)`

= `(1.8 xx 10^-3)^2/(2(9 xx 10^-6))`

= `(18 xx 10^-8)/(10^-6)`

= 0.18 J

The energy lost = 0.27 − 0.18 = 0.09 J