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

A capacitor with capacitance \(5\,\mu F\) is charged to \(5\,\mu C.\) If the plates are pulled apart to reduce the capacitance to \(2\,\mu F,\) how much work is done?

  1. A \(3.75\times 10^{-6}\,J\)
  2. B \(2.55\times 10^{-6}\,J\)
  3. C \(6.25\times 10^{-6}\,J\)
  4. D \(2.16\times 10^{-6}\,J\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(3.75\times 10^{-6}\,J\)

Step-by-step Solution

Detailed explanation

Work done \(=\Delta U\) \( = {U_t} - {U_i}\) \( = \frac{{{q^2}}}{{2{C_r}}} - \frac{{{q^2}}}{{2{C_i}}}\) \( = \frac{{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}}}{2}\left( {\frac{1}{{2 \times {{10}^{ - 6}}}} - \frac{1}{{5 \times {{10}^{ - 6}}}}} \right)\)…
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