Question

A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is _____ μ.J

Answer 1

Correct Answer: 6

The energy lost during redistribution of charge between capacitors is given by: \[ \text{Energy lost} = \frac{1}{2} \cdot \frac{C \cdot C}{C + C} \cdot (V_1 - V_2)^2, \] where \( V_1 = 200 \, \text{V} \) and \( V_2 \) is the final voltage. For two identical capacitors: \[ \text{Energy lost} = \frac{1}{2} \cdot \frac{C}{2} \cdot V_1^2. \] Substitute \( C = 600 \, \text{pF} = 600 \times 10^{-12} \, \text{F} \) and \( V_1 = 200 \, \text{V} \): \[ \text{Energy lost} = \frac{1}{2} \cdot \frac{600 \times 10^{-12}}{2} \cdot (200)^2. \] Simplify: \[ \text{Energy lost} = \frac{1}{2} \cdot 300 \times 10^{-12} \cdot 40000 = 6 \times 10^{-6} \, \text{J}. \] Convert to microjoules: \[ \text{Energy lost} = 6 \, \mu \text{J}. \] Final Answer: The energy lost is: \[ \boxed{6 \, \mu \text{J}}. \]