# A capacitor with capacitance

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?

1. (1) $6.25 \times 10^{-6} \mathrm{~J}$

2. (2) $\quad 3.75 \times 10^{-6} \mathrm{~J}$

3. (3) $2.16 \times 10^{-6} \mathrm{~J}$

4. (4) $\quad 2.55 \times 10^{-6} \mathrm{~J}$

Correct Option: 2,

Solution:

(2) $W=U_{f}-U_{i}=\frac{q^{2}}{2}\left(\frac{1}{C_{f}}-\frac{1}{C_{i}}\right)\left(\because \mathrm{U}=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}\right)$

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

$=3.75 \times 10^{-6} \mathrm{~J}$