Question

The diagram in Figure 16.74 shows three very large metal disks (seen edgewise), carrying charges as indicated. On each surface the charges are distributed approximately uniformly. Each disk has a very large radius R and a small thickness t. The distances between the disks are a and b, as shown; they also are small compared to R. Calculate ${\mathbf{V}}_{\mathbf{2}}\mathbf{-}{\mathbf{V}}_{\mathbf{1}}$, and explain your calculation briefly.

Short answer

The value of potential difference is$\frac{{\mathrm{bQ}}_{\mathrm{b}}-{\mathrm{aQ}}_{\mathrm{a}}}{{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$

Step-by-step solution

Identification of given data

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,${\mathrm{d}}_{\mathrm{b}}=\mathrm{b}$.

Concept/Significance of potential energy

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.

Determination of the potential difference.

The potential difference between the disk is given by,

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

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,

$\mathrm{E}=\frac{\mathrm{Q}}{{\mathrm{\epsilon }}_0\mathrm{A}}$

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" $$\begin{gathered} {\mathrm{V}}_2-{\mathrm{V}}_1=\frac{{\mathrm{Q}}_{\mathrm{b}}}{{\mathrm{\epsilon }}_0\mathrm{A}}\mathrm{b}-\frac{{\mathrm{Q}}_{\mathrm{a}}}{{\mathrm{\epsilon }}_0\mathrm{A}}\mathrm{a} \\ =\frac{{\mathrm{bQ}}_{\mathrm{b}}-{\mathrm{aQ}}_{\mathrm{a}}}{{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \end{gathered}$$

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