Question

A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Answer 1

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

\(E1=\frac{1}{2} CV^2=\frac{1}{2}X(600×10^-12)×(200)^2 J=1.2×10^{-5}J\)

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C_eq) of the combination is given by,

\(\frac{1}{C_eq}=\frac{1}{C}+\frac{1}{C}\)

\(\frac{1}{Ceq} =\frac{1}{600}+\frac{1}{600} = \frac{2}{600}=\frac{1}{300 }= 300 pF\)

New electrostatic energy can be calculated as

\(E2=1/2Ceq V2= \frac{1}{2}×300×(200)^2 J = 0.6 × 10-5 ^J \)

Loss in electrostatic energy = E₁ – E₂

= 1.2 × 10⁻⁵ 0.6 × 10^-5 J = 0.6 × 10^-5 J = 6 × 10^-6 J

Therefore, the electrostatic energy lost in the process is 6 × 10^-6 J .