Capacitance

Question

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

Moderate

Solution

Force of attraction between the plates of the parallel plate air capacitor is

$\mathrm{F}=\frac{{\mathrm{Q}}^{2}}{2{\mathrm{\epsilon}}_{0}\mathrm{A}}$

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

$\text{But}\mathrm{Q}=\mathrm{CV}$

$\text{and}\mathrm{C}=\frac{{\mathrm{\epsilon}}_{0}\mathrm{A}}{\mathrm{d}}\text{or}{\mathrm{\epsilon}}_{0}\mathrm{A}=\mathrm{Cd}\therefore \mathrm{F}=\frac{{\mathrm{C}}^{2}{\mathrm{V}}^{2}}{2\mathrm{Cd}}=\frac{{\mathrm{CV}}^{2}}{2\mathrm{d}}$

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