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
$(a) \frac{cv^{2}}{d} (b) \frac{c^{2}v^{2}}{2d^{2}} (c) \frac{c^{2}v^{2}}{2d} (d) \frac{cv^{2}}{2d}$
Text Solution
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Force of attraction between the plates of the parallel plate air capacitor is
$\mathrm{F} = \frac{\mathrm{Q}^2}{2 \varepsilon_0 \mathrm{A}}$ where Q is the charge on the capacitor, $\varepsilon_0$ is the permittivity of free space and A is the area of each plate. But Q = CV and $\mathrm{C} = \frac{\varepsilon_0 \mathrm{A}}{\mathrm{d}}$ or $\varepsilon_0 \mathrm{A} = \mathrm{Cd}$ $\therefore \mathrm{F} = \frac{\mathrm{C}^2 \mathrm{V}^2}{2 \mathrm{Cd}} = \frac{\mathrm{CV}^2}{2 \mathrm{d}}$
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