91Ó°ÊÓ

The given data can be listed below,

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

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,

$\mathrm{E}=\frac{(\mathrm{Q}/\mathrm{A})}{2{\mathrm{Î\mu }}_0}[1-\frac{\mathrm{z}}{{({\mathrm{R}}^2+\mathrm{z})}^{1/2}}]$ …(¾±)

$\mathrm{E}=\frac{\mathrm{Q}/\mathrm{A}}{2{\mathrm{Î\mu }}_0}[1-\frac{\mathrm{z}}{\mathrm{R}}]$ …(¾±¾±)

$E=\frac{Q/A}{2{\mathrm{Î\mu }}_0}$ …(¾±¾±¾±)

Substitute values in the equation (i)

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

Substitute all the values in the equation (ii).

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

Substitute all the values in equation (iii)

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

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.

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

Substitute values inequation (ii)

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

Substitute all the values in equation (iii)

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

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.