Figure 22-58ashows a circular disk that is uniformly charged. The central zaxis is perpendicular to the disk face, with the origin at the disk. Figure 22-58bgives the magnitude of the electric field along that axis in terms of the maximum magnitude $E_m$at the disk surface. The zaxis scale is set by${\mathit{z}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{.0}\mathbf{}\mathbf{\text{cm}}$. What is the radius of the disk?

• A circular disk of uniform charge.
• The central z-axis is perpendicular to the face of the disk.
• Maximum magnitude of the electric field is.$E_m$
• The z-axis scale set is.$z_s=8\text{\hspace{0.17em}cm}$

Using the concept of the electric field at a point from the disk along its central axis, we can get the ratio of the electric field at a point and maximum electric field, thus solving it by substituting the given data, we can get the radius of the disk.

Formula:

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

where,= surface charge density

z = distanceon the central axis of the disk

R = Radius of the disk

Using equation (i) for the electric field at a point from the disk and maximum electric field (at$z=0$), we can get that

$\begin{array}{c}\frac{E}{E_{max}}=1-\frac{z}{\sqrt{z^2+R^2}}\\ \frac{1}{2}=1-\frac{z}{\sqrt{z^2+R^2}}\text{(}\because \text{fromgraph,}\frac{E}{E_{max}}=\frac{1}{2}\text{)}\\ \frac{z}{\sqrt{z^2+R^2}}=\frac{1}{2}\\ 3z^2=R^2\\ R=z\sqrt{3}\\ =4\times \sqrt{3}\text{\hspace{0.17em}cm}(\because z=4,fromgraph)\\ =6.93\text{\hspace{0.17em}cm}\end{array}$

Hence, the value of the radius of the disk is .$6.93\text{\hspace{0.17em}cm}$