Question

A disk of radius 16 cm has a total charge 4 × 10−6 C distributed uniformly all over the disk. (a) Using the exact equation, what is the electric field 1 mm from the center of the disk? (b) Using the same exact equation, find the electric field 3 mm from the center of the disk. (c) What is the percent difference between these two numbers?

Short answer

a) The electric field 1 mm from the center of the disk is$2.79\times {10}^6\mathrm{N}/\mathrm{C}$

b) The electric field 3 mm from the center of the disk is $2.75\times {10}^6\mathrm{N}/\mathrm{C}$.

c) 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 1 mm.

Step-by-step solution

Identification of given data

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,$\mathrm{q}=4\times {10}^{-6}\mathrm{C}$

Concept/Significance of coulomb law

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.

(a) Determination of the electric field 1 mm from the center of the disk

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

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

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$1\mathrm{mm}\left(1-\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=1\times {10}^{-3}\mathrm{m}$ and R is the radius of disk.

Substitute all the values in the above expression.

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

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

(b) Determination of the electric field 3 mm from the center of the disk

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

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

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 $3\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=3\times {10}^{-3}\mathrm{m}$and R is the radius of disk.

Substitute all the values in the above expression.

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

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

(b) Determination of the percent difference between these two numbers of electric field

The percentage change formula is given by,

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

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

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

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.