91Ó°ÊÓ

The given data can be listed below,

• The radius of the disk is,$\mathrm{R}=16\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$
• The total charge on the disk is,$\mathrm{q}=4×{10}^{-6}\mathrm{C}$

Coulomb's law is a fundamental concept in physics about electricity. The law examines the forces that generate when two charged objects collide. Attractions and electrostatic fields decrease with increasing distance.

The magnitude of the electric field along the axis of disk is given by,

$\mathrm{E}=\frac{\mathrm{q}/\mathrm{A}}{2{\mathrm{Î\mu }}_0}\left(1-\frac{\mathrm{r}}{\sqrt{{\mathrm{R}}^2+{\mathrm{r}}^2}}\right)$

Here, q is the charge on the disk, A is the area of disk, r is the radial distance from the centre of disk whose value is$1\mathrm{mm}\left(1-\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=1×{10}^{-3}\mathrm{m}$ and R is the radius of disk.

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{E}=\frac{4×{10}^{-6}\mathrm{C}}{2\mathrm{Ï€}{\left(0.21\mathrm{m}\right)}^2{\mathrm{Î\mu }}_0}\left(\frac{1×{10}^{-3}\mathrm{m}}{\sqrt{{\left(0.16\mathrm{m}\right)}^2+{\left({10}^{-3}\mathrm{m}\right)}^2}}\right) \\ =2.79×{10}^6\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the electric field 1 mm from the center of the disk is$2.79×{10}^6\mathrm{N}/\mathrm{C}$

The magnitude of the electric field along the axis of disk is given by,

$\mathrm{E}=\frac{\mathrm{q}/\mathrm{A}}{2{\mathrm{Î\mu }}_0}\left(1-\frac{\mathrm{r}}{\sqrt{{\mathrm{R}}^2+{\mathrm{r}}^2}}\right)$

Here, q is the charge on the disk, A is the area of disk, r is the radial distance from the centre of disk whose value is $3\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=3×{10}^{-3}\mathrm{m}$and R is the radius of disk.

Substitute all the values in the above expression.

$$\begin{gathered} \mathrm{E}=\frac{4×{10}^{-6}\mathrm{C}}{2\mathrm{Ï€}{\left(0.21\mathrm{m}\right)}^2{\mathrm{Î\mu }}_0}\left(1-\frac{3×{10}^{-3}\mathrm{m}}{\sqrt{{\left(0.16\mathrm{m}\right)}^2+{\left({10}^{-3}\mathrm{m}\right)}^2}}\right) \\ =2.75×{10}^6\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the electric field 3 mm from the center of the disk is$2.75×{10}^6\mathrm{N}/\mathrm{C}$ .

The percentage change formula is given by,

$\%=\frac{\mathrm{original}\mathrm{value}-\mathrm{value}\mathrm{after}\mathrm{change}}{\mathrm{original}\mathrm{value}}×100$

So, the percentage change for electric field is given by,

$$\begin{gathered} \frac{{\mathrm{E}}_1{\mathrm{E}}_3}{{\mathrm{E}}_1}×100=\left(\frac{2.79×{10}^6\mathrm{N}/\mathrm{C}-2.75×{10}^6\mathrm{N}/\mathrm{C}}{2.79×{10}^6\mathrm{N}/\mathrm{C}}\right)×100 \\ =1.43\% \end{gathered}$$

Thus, the value of percentage change is -1.43% which means the electric field at the distance of 3 mm is decrease by 1.43% compared to the electric field at the distance of 1mm.