First slide

Potential energy

Question

The force of attraction between the plates of air filled capacitor having charge Q and area of each plate A is given by

Moderate

A

$\frac{Q^{2}}{2 ε_{0} A}$
B

$\frac{Q}{2 ε_{0} A}$
C

$\frac{Q^{2}}{ε_{0} A}$
D

$\frac{Q}{2 ε_{0} A^{2}}$

Solution

Let x be the separation between the plates of the condenser. The energy stored in the capacitor is given by
$U = \frac{1}{2} {CV}^{2} = \frac{ε_{0} {AV}^{2}}{2 x} (∵ C = \frac{ε_{0} A}{x}) ∴ \frac{dU}{dx} = - \frac{ε_{0} {AV}^{2}}{2 x^{2}}$
So, $F = - \frac{dU}{dx} = \frac{ε_{0} {AV}^{2}}{2 x^{2}} (1)$
Further, $Q = CV = \frac{ε_{0} AV}{x}$
or $V = \frac{Qx}{ε_{0} A} . . . (2)$
From eqs. (1) and (2), we get
$F = \frac{ε_{0} A}{2 x^{2}} \times {(\frac{Qx}{ε_{0} A})}^{2} = \frac{Q^{2}}{2 ε_{0} A}$

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