Question:
A capacitor with capacitance $5 \mu \mathrm{F}$ is charged to $5 \mu \mathrm{C}$. If the plates are pulled apart to reduce the capacitance to $2 \mu \mathrm{F}$, how much work is done?
Correct Option: 2,
Solution:
(2) $W=U_{f}-U_{i}=\frac{q^{2}}{2}\left(\frac{1}{C_{f}}-\frac{1}{C_{i}}\right)\left(\because \mathrm{U}=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}\right)$
$=\frac{\left(5 \times 10^{-6}\right)^{2}}{2}\left(\frac{1}{2}-\frac{1}{5}\right) \times 10^{6}$
$=3.75 \times 10^{-6} \mathrm{~J}$
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