91Ó°ÊÓ

The distance between the capacitor disks plates is $s$.

The thickness of the disk is $t$.

The radius of the disk is $R$.

The net charge on left disk is $+Q$and on right disk is $-Q$.

The expression to calculate the electrical field between the places is given as follows.

$\mathit{E}\mathbf{=}\frac{\mathbf{Q}}{{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{A}}$ …… (i)

Here,$A$ is the area of the disk.

The expression to calculate the potential difference between the point 1 and 2 is given as follows.

${\mathit{V}}_{\mathbf{2}}\mathbf{-}{\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathit{E}\mathit{d}$…… (ii)

The distance between the point 1 and 2 is $s/2$.

The area of the disk is given by,

$A=\mathrm{Ï€}{R}^2$

Calculate the electrical field.

Substitute $\mathrm{Ï€}{R}^2$for $A$into equation (i).

$E=\frac{Q}{{\mathrm{Î\mu }}_0\mathrm{Ï€}{R}^2}$

Calculate the potential difference between the point 1 and 2,

Substitute $\frac{Q}{{\mathrm{Î\mu }}_0\mathrm{Ï€}{R}^2}$for $E$and $s/2$for $d$into equation (ii).

$$\begin{gathered} V_2-V_1=\frac{Q}{{\mathrm{Î\mu }}_0\mathrm{Ï€}{R}^2}×\frac{s}{2} \\ {V}_2-V_1=\frac{Qs}{2{\mathrm{Î\mu }}_0\mathrm{Ï€}{R}^2} \end{gathered}$$

Hence the potential difference between the points 1 and 2 is$\frac{Qs}{2{\mathrm{Î\mu }}_0\mathrm{Ï€}{R}^2}$ .