CUET · PHYSICS · PYQ PAPER 2025
A 400 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 400 pF capacitor. The electrostatic energy lost in the process is
- A \( 2.4 \times 10^{-6} \text{ J} \)
- B \( 4.\times 10^{-6} \text{ J} \)
- C \( 6.0 \times 10^{-6} \text{ J} \)
- D \( 8.0 \times 10^{-6} \text{ J} \)
Answer & Solution
Correct Answer
(B) \( 4.\times 10^{-6} \text{ J} \)
Step-by-step Solution
Detailed explanation
\(U_{lost} = \frac{1}{4} C V^2\) \(U_{lost} = \frac{1}{4} (400 \times 10^{-12} \text{ F})(200 \text{ V})^2\) \(U_{lost} = 4 \times 10^{-6} \text{ J}\)
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