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?
Pick an option to check your answer.
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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