Suppose that the radius of a disk is 21 cm, and the total charge distributed uniformly all over the disk is $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{C}$. (a) Use the exact result to calculate the electric field 1 mm from the center of the disk. (b) Use the exact result to calculate the electric field 3 mm from the center of the disk. (c) Does the field decrease significantly?

The given data can be listed below,

• The radius of the disk is,$R=21cm$
• Total charge on the disk is,$q=5\times {10}^{-6}C$

An electric field is emitted by all charges. Because a positive charge has a positive electric field, the positive direction is defined as outward pointing and the electric fields of negative charges are inward-pointing.

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

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

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 localid="1656936761969" $1mm\left(\frac{{10}^{-3}m}{1mm}\right)=1\times {10}^{-3}m$and R is the radius of disk.

Substitute all the values in the above expression.

$$\begin{gathered} E=\frac{5\times {10}^{-6}C}{2\pi {\left(0.21m\right)}^2{\epsilon }_0}\left(1-\frac{1\times {10}^{-3}m}{\sqrt{{\left(0.21m\right)}^2+{\left({10}^{-3}m\right)}^2}}\right) \\ =2.029\times {10}^{6}N/C \end{gathered}$$

Thus, the electric field 1 mm from the center of the disk is $2.029\times {10}^{6}N/C$.

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

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

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

Substitute all the values in the above expression.

$$\begin{gathered} E=\frac{5\times {10}^{-6}C}{2\pi {\left(0.21m\right)}^2{\epsilon }_0}\left(1-\frac{3\times {10}^{-3}m}{\sqrt{{\left(0.21m\right)}^2+{\left(3\times {10}^{-3}m\right)}^2}}\right) \\ =2.01\times {10}^{6}N/C \end{gathered}$$

Thus, the electric field 3 mm from the center of the disk$2.01\times {10}^{6}N/C$.

The magnitude of both the electric fields at 1 mm and 3 mm have only almost 1% difference and radius of disk is very greater than radial distances so it does not effect the electric field. Electric field in this case can be written as,

$E=\frac{q}{2A{\epsilon }_0}$

The above expression gives a constant value.

Thus, the electric field does not decrease.