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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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 \(nJ\))

  1. A \(5\)
  2. B \(6\)
  3. C \(7\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(6\)

Step-by-step Solution

Detailed explanation

\(\Delta \mathrm{Q}_{\mathrm{L}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}-\left[\frac{(\mathrm{Q} / 2)^{2}}{2 \mathrm{C}} \times 2\right]=\frac{\mathrm{Q}^{2}}{4 \mathrm{C}}\) \(=\frac{1}{4} \mathrm{CV}^{2}\) \(=\frac{1}{4} \times 60 \times 10^{-12} \times 4 \times 10^{2}\)…
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