JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2020
- Medium
- Numerical
A \(60\; pF\) capacitor is fully charged by a \(20\; \mathrm{V}\) supply. It is then disconnected from the supply and is connected to another uncharged \(60 \;pF\) capactior is parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in \(nJ\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{Q}_{\mathrm{L}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}-\left[\frac{(\mathrm{Q} / 2)^{2}}{2 \mathrm{C}} \times 2\right]=\frac{\mathrm{Q}^{2}}{4 \mathrm{C}}\)
\(=\frac{1}{4} \mathrm{CV}^{2}\)
\(=\frac{1}{4} \times 60 \times 10^{-12} \times 4 \times 10^{2}\)
\(=6 \mathrm{nJ}\)
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