For a disk of radius R=20cm and $\mathbf{Q}\mathbf{=}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{C}$, calculate the electric field 2 mm from the center of the disk using all three equations:

role="math" localid="1656928965291" $\mathbf{E}\mathbf{=}\frac{(Q/A)}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{[}\mathbf{1}\mathbf{-}\frac{\mathbf{z}}{{\left(R^2+z\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}}\mathbf{]}$

$\mathbf{E}\mathbf{\approx }\frac{\mathbf{Q}\mathbf{/}\mathbf{A}}{\mathbf{2}{\mathbf{e}}_{\mathbf{0}}}\mathbf{[}\mathbf{1}\mathbf{-}\frac{\mathbf{z}}{\mathbf{R}}\mathbf{]}\mathbf{,}\mathbf{and}\mathbf{}\mathbf{E}\mathbf{\approx }\frac{\mathbf{Q}\mathbf{/}\mathbf{A}}{\mathbf{2}{\mathbf{e}}_{\mathbf{0}}}$

How good are the approximate equations at this distance? For the same disk, calculate E at a distance of 5 cm (50 mm) using all three equations. How good are the approximate equations at this distance?

The given data can be listed below,

• The radius of the disk is,R=20cm
• The charge on the disk is,$\mathrm{Q}=6\times {1010}^{-6}\mathrm{C}$

An electric field is a mathematical construct that represents the amount and direction of the net electrical force experienced by a unit of electrical charge at a particular place in space as a result of interaction with all other electrical charges in the area.

The equations for electric field are given by,

$\mathrm{E}=\frac{(\mathrm{Q}/\mathrm{A})}{2{\mathrm{\epsilon }}_0}[1-\frac{\mathrm{z}}{{({\mathrm{R}}^2+\mathrm{z})}^{1/2}}]$ …(i)

$\mathrm{E}=\frac{\mathrm{Q}/\mathrm{A}}{2{\epsilon }_0}[1-\frac{\mathrm{z}}{\mathrm{R}}]$ …(ii)

$E=\frac{Q/A}{2{\epsilon }_0}$ …(iii)

Substitute values in the equation (i)

$$\begin{gathered} \mathrm{E}=\frac{(6\times {10}^{-6}\mathrm{C}/\mathrm{\pi }{\left(0.20\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left(1-\frac{2\times {10}^{-3}\mathrm{m}}{{\left(0.20\mathrm{m}\right)}^2+{\left(2\times {10}^{-3}\mathrm{m}\right)}^2}\right) \\ =0.027\times {10}^{10}\left(0.99\right)\mathrm{N}/\mathrm{C} \\ =2.67\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$

Substitute all the values in the equation (ii).

$$\begin{gathered} \mathrm{E}=\frac{0.477\times {10}^{-2}C/m}{17.7\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left(1-\frac{2\times {10}^{-3}\mathrm{m}}{\left(0.20\mathrm{m}\right)}\right) \\ =2.66\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$

Substitute all the values in equation (iii)

$$\begin{gathered} \mathrm{E}=\frac{(6\times {10}^{-6}\mathrm{C}/\mathrm{\pi }{\left(0.20\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2} \\ =\frac{0.477\times {10}^{-2}C/m}{17.7\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2} \\ =2.69\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$

The approximate equations give the accurate and same answers for 2 mm distance

Substitute values in equation (i), (ii), (iii) for the electric field at a distance of 5 cm.

$$\begin{gathered} \mathrm{E}=\frac{(6\times {10}^{-6}\mathrm{C}/\mathrm{\pi }{\left(0.20\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left(1-\frac{5\times {10}^{-2}\mathrm{m}}{\sqrt{{\left(0.20\mathrm{m}\right)}^2+{\left(5\times {10}^{-3}\mathrm{m}\right)}^2}}\right) \\ =0.027\times {10}^{10}\left(0.75\right)\mathrm{N}/\mathrm{C} \\ =2.04\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$

Substitute values inequation (ii)

$$\begin{gathered} \mathrm{E}=\frac{0.477\times {10}^{-2}\mathrm{C}/\mathrm{m}}{17.7\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left(1-\frac{5\times {10}^{-3}\mathrm{m}}{0.20\mathrm{m}}\right) \\ =2.02\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$

Substitute all the values in equation (iii)

$$\begin{gathered} \mathrm{E}=\frac{(6\times {10}^{-6}\mathrm{C}/\mathrm{\pi }{\left(0.20\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2} \\ =\frac{0.477\times {10}^{-2}C/m}{17.7\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2} \\ =2.69\times {10}^8\mathrm{N}/\mathrm{C} \end{gathered}$$

The approximate equations do not give same answers for 5 mm distance

Thus, the approximate equations for the electric field only give an accurate answer for the distance nearer to the centre of the disk, that is, for 1 mm and a less precise solution for the distance away from the centre, i.e., 5 cm.