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

• A. $\frac{CV^2}{d}$
• B. $\frac{C^2V^2}{2d^2}$
• C. $\frac{C^2V^2}{2d}$
• D. $\frac{CV^2}{2d}$

Answer 1

Answer: (D)

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

Answer 2

Force of attraction between the plates of the parallel plate air capacitor is
$F= \frac{Q^2}{2 \varepsilon_{0}A}$
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
$C = \frac{\varepsilon_{0}A}{d}$ or $\varepsilon_{0}A = Cd$
$\therefore F = \frac{C^2V^2}{2Cd} = \frac{CV^2}{2d}$