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 3µF 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 electrostatic energy in the capacitor

= \(\frac{1}{2}\)Cv² = \(\frac{1}{2}\)(6 × 10^-6)(300)²

= 3 × 10^-6 × 9 × 10⁴ = 0.27J

The charge on this capacitor,

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^-6F

By conservation of charge, Q = 1.8 C.

∴ The energy of the system = \(\frac{Q^{2}}{2 C}\)

= \(\frac{\left(1.8 \times 10^{-3}\right)^{2}}{2\left(9 \times 10^{-6}\right)}=\frac{18 \times 10^{-8}}{10^{-6}}\) = 0.18 J

The energy lost = 0.27 – 0.18 = 0.09 J