91Ó°ÊÓ

The given data can be listed below,

• ${\mathrm{V}}_2-{\mathrm{V}}_1$The radius of the disk is, R .
• The thickness of the disk is, t .
• The distance between upper and middle disk is, ${\mathrm{d}}_{\mathrm{a}}=\mathrm{a}$.
• The distance between middle and bottom disk is,${\mathrm{d}}_{\mathrm{b}}=\mathrm{b}$.

The force created in both the positive and negative ends of the electrical energy source is known as potential energy.

Which causes the movement of positively and negatively charged particles in a conductor to create or convert electrical energy into a usable form.

The potential difference between the disk is given by,

${\mathrm{V}}_2-{\mathrm{V}}_1={\mathrm{E}}_{\mathrm{b}}{\mathrm{d}}_{\mathrm{b}}-{\mathrm{E}}_{\mathrm{a}}{\mathrm{d}}_{\mathrm{a}}…\left(1\right)$

Here, $E_b$is the electric field between middle and bottom disk, $E_a$is the electric field between middle and upper disk, $d_b$is the distance between middle and bottom disk, and $d_a$distance between upper and middle disk.

The electric field inside the capacitor is given by,

$\mathrm{E}=\frac{\mathrm{Q}}{{\mathrm{Î\mu }}_0\mathrm{A}}$

Here, Q is the charge in the plate, and A is the area of the plate.

Substitute the value of electric field in equation (1).

role="math" localid="1657087519387" $$\begin{gathered} {\mathrm{V}}_2-{\mathrm{V}}_1=\frac{{\mathrm{Q}}_{\mathrm{b}}}{{\mathrm{Î\mu }}_0\mathrm{A}}\mathrm{b}-\frac{{\mathrm{Q}}_{\mathrm{a}}}{{\mathrm{Î\mu }}_0\mathrm{A}}\mathrm{a} \\ =\frac{{\mathrm{bQ}}_{\mathrm{b}}-{\mathrm{aQ}}_{\mathrm{a}}}{{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2} \end{gathered}$$

Thus, the value of potential difference is role="math" localid="1657087483598" $\frac{{\mathrm{bQ}}_{\mathrm{b}}-{\mathrm{aQ}}_{\mathrm{a}}}{{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}$