Question

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?

Short answer

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.

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of electric field

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.

Determination of the electric field 2 mm from the center of the disk using all three equations

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

Step 4: Determination of the electric field at a distance of 5 cm (50 mm) using all three equations

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.