JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A capacitor of capacitance \(50 \; pF\) is charged by \(100 \; V\) source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is \(\dots \; nJ\).
- A \(155\)
- B \(145\)
- C \(135\)
- D \(125\)
Answer & Solution
Correct Answer
(D) \(125\)
Step-by-step Solution
Detailed explanation
Energy loss \(=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}\) \(=\frac{1}{2} \frac{50 \times 50 \times 10^{-12} \times 10^{-12}}{(50+50) 10^{-12}}(100-0)^{2}=125 \; n J\)
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