A parallel plate air capacitor has capacity C, distance of separa…

MCQ

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
$\frac{{{C^2}{V^2}}}{{2{d^2}}}$
B
$\;\frac{{{C^2}{V^2}}}{{2d}}$
✓
$\frac{{C{V^2}}}{{2d}}$
D
$\;\frac{{C{V^2}}}{d}$

✓

Answer

Correct option: C.

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

c
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 freespace and $A$ is the area of each plate.

But $Q=C V$

and $C=\frac{\varepsilon_{0} A}{d}$ or $\varepsilon_{0} A=C d \quad \therefore \quad F=\frac{C^{2} V^{2}}{2 C d}=\frac{C V^{2}}{2 d}$

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