Question

A parallel plate capacitor has plate of area $A$ and separation $d$. It is charged to a potential difference $V_{0}$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

• A. $\frac{\varepsilon_{0} A V_{0}^{2}}{3 d}$
• B. $\frac{\varepsilon_{0} A V_{0}^{2}}{2 d}$
• C. $\frac{\varepsilon_{0} A V_{0}^{2}}{4 d}$
• D. $\frac{\varepsilon_{0} A V_{0}^{2}}{d}$

Answer 1

Answer: (D)

$C=\frac{\varepsilon_{0} A}{d}$ and charge $Q=C V_{0}$
When spacing is made $(3 d)$
$C'=\frac{\varepsilon_{0} A}{3 d}=\frac{C}{3}$
As battery is disconnected, charge $Q$ is constant.
$\therefore$ Work done $=$ Final energy $-$ Initial energy
$=\frac{Q^{2}}{2 C'}-\frac{Q^{2}}{2 C}$
$=\frac{Q^{2} \times 3}{2 C}-\frac{Q^{2}}{2 C}$
$=\frac{2 \times Q^{2}}{2 C}=\frac{\left(C V_{0}\right)^{2}}{C}$
$=C V_{0}^{2}=\frac{\varepsilon_{0} A V_{0}^{2}}{d}$