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?
JEE MAIN 2019, Medium
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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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