A parallel plate air capacitor has capacity , distance of separation between plates is , and potential difference is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is [2015]

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Solution

Q.
A parallel plate air capacitor has capacity $C$ , distance of separation between plates is $d$ , and potential difference $V$ is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is [2015]

1 $\frac{C V^{2}}{d}$

2 $\frac{C^{2} V^{2}}{2 d^{2}}$

3 $\frac{C^{2} V^{2}}{2 d}$

4 $\frac{C V^{2}}{2 d}$

Ans.
(4)

Force of attraction between the plates of the parallel plate air capacitor is $F = \frac{Q^{2}}{2 ε_{0} A}$

where $Q$ is the charge on the capacitor, $ε_{0}$ is the permittivity of free space, and $A$ is the area of each plate.

But $Q = C V$ and $C = \frac{ε_{0} A}{d}$ or $ε_{0} A = C d$

$∴$ $F = \frac{C^{2} V^{2}}{2 C d} = \frac{C V^{2}}{2 d}$

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