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Physics · Electrostatic Potential
हिन्दीA \( 600\text{ pF} \) capacitor is charged by a \( 200\text{ V} \) supply. It is then disconnected from the supply and connected to another uncharged \( 600\text{ pF} \) capacitor in parallel. The electrostatic energy lost as heat and electromagnetic radiation in the process is:
Correct answer · Option 4
\( 6.0 \times 10^{-6}\text{ J} \)
Explanation
Loss of energy \( \Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 = \frac{1}{2} \frac{600 \times 600}{1200} \times 10^{-12} \times (200 - 0)^2 = \frac{1}{4} \times 600 \times 10^{-12} \times 40000 = 6.0 \times 10^{-6}\text{ J} \).
