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. $\dfrac{{{C^2}{V^2}}}{{2{d^2}}}$
B. $\dfrac{{{C^2}{V^2}}}{{2d}}$
C. $\dfrac{{C{V^2}}}{{2d}}$
D. $\dfrac{{{C^2}V}}{d}$

Answer 1

To find the force between the plates find the electric field by one of the plates (out of those two plates of the capacitor) nearby other plate and then multiply the charge of the this plate with the electric field calculated you will get the required force.

Complete step by step answer:
Given potential across capacitor is $V$ and capacitance is $C$. So using $Q = CV$ magnitude of charge on each plate will be $CV$. Now assuming ideal conditions we will find an electric field by one of the plates. For ideal conditions we can treat the plates as an infinitely charged sheet (as separation between the plates is very small). And we know that electric field due to infinite sheet is,
$E = \dfrac{\sigma }{{2{\varepsilon _o}}}$ where, $\sigma$ is areal charge density,
Therefore,
$\sigma = \dfrac{Q}{A}\\\ \Rightarrow E = \dfrac{Q}{{2A{\varepsilon _o}}} \\\ \Rightarrow E= \dfrac{{CV}}{{2A{\varepsilon _o}}}$
Now using $F = QE$
We have
$F = QE\\\ \Rightarrow F = \dfrac{{{Q^2}}}{{2A{\varepsilon _o}}}\\\ \Rightarrow F = \dfrac{{{C^2}{V^2}}}{{2A{\varepsilon}}}$
Also $C = \dfrac{{A{\varepsilon _o}}}{d}$
Putting value of C we have,
$\therefore F = \dfrac{{C{V^2}}}{{2d}}$

Therefore, Option C is correct.

Additional information:
Parallel Plate Capacitors are formed by an arrangement of electrodes and insulating material or dielectric. A parallel plate capacitor can only store a finite amount of energy before dielectric breakdown occurs. It can be defined as: when two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor.

Note: Here finding electric field due to one of the charge on the other plate is crucial once you calculate the value of electric field then you can simply multiply electric field with the charge to get total force also since we assume ideal cases therefore we can treat individual plates as infinite charged sheets for distances very close to the plates.