Question

A capacitor with capacitance $5 \mu F$ is charged to $5 \mu C$. If the plates are pulled apart to reduce the capacitance to $2$ $1/4 F$, how much work is done ?

• A. 3.75 × 10$^{-6}$ J
• B. 2.55 × 10$^{-6}$ J
• C. 2.16 × 10$^{-6}$ J
• D. 6.25 × 10$^{-6}$ J

Answer 1

Answer: (A)

Work done = $\Delta$U
$ \, \, \, \, \, \, \, \, \, \, \, \, \, = U_f - U_i$
$\, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{q^2}{2C_r} - \frac{q^2}{2C_i}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{(5 \times 10^{-6})^2}{2} \bigg(\frac{1}{2 \times 10^{-6}} - \frac{1}{5 \times 10^{-6}}\bigg)$
$\, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{15}{4} \times 10^{-6}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, = 3.75 \times 10^{-6}J$