Question

At what distance along the central perpendicular axis of a uniformly charged plastic disk of radius$\mathbf{0}\mathbf{.600}\mathbf{}\mathbf{\text{m}}$is the magnitude of the electric field equal to one-half the magnitude of the field at the center of the surface of the disk?

Short answer

The distance value along the central perpendicular axis of a uniformly charged plastic disk where the magnitude of electric field is one-half the magnitude of the field at the center of the surface of the disk is.$0.346\text{\hspace{0.17em}m}$

Step-by-step solution

The given data

• Radius of the disk,$R=0.6\text{\hspace{0.17em}m}$
• The magnitude of field at the point is one-half the magnitude of the field at the center of the disk,$E/E_c=1/2.$

Understanding the concept of electric field 

Using the given formula of the electric field at a point due to the disk and at the center of the disk, we can get the relation of the distance to that to the radius of the disk. This will determine the value of the distance.

Formula:

The magnitude of the electric field produced by the disk at a point on its central axis, $E=\frac{\sigma }{2{\epsilon }_o}(1-\frac{z}{\sqrt{z^2+R^2}})$ (i)

where,= surface charge density

z = distanceon the central axis of the disk

R = Radius of the disk

Calculation of the distance of the point

The magnitude of the field at the center of the disk (z = 0) is given using equation (i) as:

$E_c=\sigma /2{\epsilon }_o$…… (a)

From the given data,$E/E_c=1/2.$

Now, substituting the value of electric fields from equation (i) and equation (a) and using it in the above equation, we can get the distance value as follows:

$\begin{array}{c}\frac{\sigma }{2{\epsilon }_o}(1-\frac{z}{\sqrt{z^2+R^2}})/\frac{\sigma }{2{\epsilon }_o}=1/2\\ 1-\frac{z}{\sqrt{z^2+R^2}}=\frac{1}{2}\\ \frac{z}{\sqrt{z^2+R^2}}=\frac{1}{2}\\ 3z^2=R^2\\ z=R/\sqrt{3}\\ =0.6\text{\hspace{0.17em}m}/\sqrt{3}\\ =0.346\text{m}\end{array}$

Hence, the value of the distance for this case is.$0.346\text{\hspace{0.17em}m}$