Question

Obtain an expression for the capacitance of a parallel plate capacitor with air between the plates.

Answer 1

Consider a parallel plate capacitor consisting of two large conducing plates held parallel to each other and separated by a small distance (d), ($A$$>$$>$ d${}^2$),

A $\to$ area of each plate

if plate (1) carries a charge + Q and (2) carries - Q then, due to attraction the charges exist only on inner surfaces facing each other.

Let surface density of charge on (1) and (2) be ${\sigma }_1=\frac{Q}{A}$ and ${\sigma }_2=\frac{Q}{A}$ respectively.

The electric field due to a plane charged sheet, at a point close to it in vacuum is $\frac{\sigma }{2{ϵ}_0}$directed normal to the surface.

The field at a point P between the plates then is:

$E$= $E_1$ + $E_2$ =$\frac{\sigma }{2{ϵ}_0}$ + $\frac{\sigma }{2{ϵ}_0}$ = $\frac{\sigma }{{ϵ}_0}\to$(1)

along the direction from (1) to (2), due to both plates.

The work done in moving a unit positive charge from the negative to positive plate against field is $E$d. Which is by definition `V' or the potential difference between the plates.

Hence, we have V = $E$d = $\frac{Qd}{{ϵ}_{0}A}\to$(2)

$\therefore$ Capacitance of the parallel plate capacitor is $C=\frac{Q}{V}=\frac{{ϵ}_{0}A}{d}\to$(3)