91Ó°ÊÓ

• 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.$z_s=8\text{ c³¾}$

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, $E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o}\left(1−\frac{z}{\sqrt{z^2+R^2}}\right)$ (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

$\begin{array}{c}\frac{E}{E_{max}}=1−\frac{z}{\sqrt{z^2+R^2}}\\ \frac{1}{2}=1−\frac{z}{\sqrt{z^2+R^2}}\text{(}âˆ\mu \text{fromgraph,}\frac{E}{E_{max}}=\frac{1}{2}\text{)}\\ \frac{z}{\sqrt{z^2+R^2}}=\frac{1}{2}\\ 3z^2=R^2\\ R=z\sqrt{3}\\ =4×\sqrt{3}\text{ c³¾}(âˆ\mu z=4,fromgraph)\\ =6.93\text{ c³¾}\end{array}$

Hence, the value of the radius of the disk is .$6.93\text{ c³¾}$