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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:
A
$\frac{C^2V^2}{2d^2}$
B
$\frac{C^2V^2}{2d}$
C
$\frac{CV^2}{2d}$
D
$\frac{CV^2}{d}$
Detailed Solution
$F = \frac{Q^2}{2\varepsilon_0A}$
$Q = CV$ and $C = \frac{\varepsilon_0A}{d} \Rightarrow \varepsilon_0A = Cd$
$F = \frac{C^2V^2}{2Cd} = \frac{CV^2}{2d}$
$Q = CV$ and $C = \frac{\varepsilon_0A}{d} \Rightarrow \varepsilon_0A = Cd$
$F = \frac{C^2V^2}{2Cd} = \frac{CV^2}{2d}$