First slide

Capacitance

Question

A parallel plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their separation is 2d. What is the work required to separate the plates?

Moderate

A

$2 ε_{0} A V^{2} / d$
B

$ε_{0} A V^{2} / d$
C

$3 ε_{0} A V^{2} / 2 d$
D

$ε_{0} A V^{2} / 2 d$

Solution

$W = U_{2} - U_{1} = \frac{q^{2}}{2} [\frac{1}{C_{2}} - \frac{1}{C_{1}}]$
$C_{1} = \frac{ε_{0} A}{d}, C_{2} = \frac{C_{1}}{2} = \frac{ε_{0} A}{2 d}$
$q = C_{1} V = \frac{ε_{0} A V}{d}$ $\Rightarrow W = \frac{1}{2} \frac{ε_{0} A V^{2}}{d}$
Alternatively,
$W = F d = \frac{Q^{2}}{2 A ε_{0}} d = \frac{C_{1}^{2} V^{2}}{2 ε_{0} A} d = \frac{1}{2} \frac{ε_{0} A V^{2}}{d}$

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