First slide

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

A

$\frac{{CV}^{2}}{d}$
B

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

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

$\frac{{CV}^{2}}{2 d}$

Solution

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.
$But Q = CV$
$and C = \frac{ε_{0} A}{d} or ε_{0} A = Cd ∴ F = \frac{C^{2} V^{2}}{2 Cd} = \frac{{CV}^{2}}{2 d}$

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