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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2019
  • Medium
  • MCQ

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?

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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Work done \(=\Delta U\)

\( = {U_t} - {U_i}\)

\( = \frac{{{q^2}}}{{2{C_r}}} - \frac{{{q^2}}}{{2{C_i}}}\)

\( = \frac{{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}}}{2}\left( {\frac{1}{{2 \times {{10}^{ - 6}}}} - \frac{1}{{5 \times {{10}^{ - 6}}}}} \right)\)

\( = \frac{{15}}{4} \times {10^{ - 6}}\)

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

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