JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 13)
A capacitor with capacitance 5μF is charged to
5μC. If the plates are pulled apart to reduce the
capacitance to 2μF, how much work is done ?
2.16 × 10–6 J
2.55 × 10–6 J
3.75 × 10–6 J
6.25 × 10–6 J
Explanation
Work done = $$\Delta $$U
= Uf – Ui
$$ = {{{q^2}} \over {2{C_r}}} - {{{q^2}} \over {2{C_i}}}$$
$$ = {{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}} \over 2}.\left( {{1 \over {2 \times {{10}^{ - 6}}}} - {1 \over {5 \times {{10}^{ - 6}}}}} \right)$$
$$ = {{15} \over 4} \times {10^{ - 6}} = 3.75{\rm{ }} \times {\rm{ }}{10^{-6}}{\rm{ }}J$$
= Uf – Ui
$$ = {{{q^2}} \over {2{C_r}}} - {{{q^2}} \over {2{C_i}}}$$
$$ = {{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}} \over 2}.\left( {{1 \over {2 \times {{10}^{ - 6}}}} - {1 \over {5 \times {{10}^{ - 6}}}}} \right)$$
$$ = {{15} \over 4} \times {10^{ - 6}} = 3.75{\rm{ }} \times {\rm{ }}{10^{-6}}{\rm{ }}J$$
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