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  4. 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?

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Q. 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?

Electrostatic Potential and Capacitance

Solution:

$W= U_2 - U_1 = \frac{q^2}{2}[\frac{1}{C_2} -\frac{1}{C_1}]$
$C_1 = \frac{\varepsilon_0A}{d} , C_2 = \frac{C_1}{2} = \frac{\varepsilon_0A}{2d}$
$q = C_1V = \frac{\varepsilon_0AV}{d}$
Solve to get $W = \frac{1}{2} \frac{\varepsilon_0AV^2}{d}$
Alternatively:
$W = Fd = \frac{Q^2}{2A\varepsilon_0} d $
$ = \frac{C^2_1V^2}{2\varepsilon_0A} d = \frac{1}{2} \frac{\varepsilon_0AV^2}{d}$