Question

The quarter disc of radius $R$ (see figure) has a uniform surface charge density $\sigma$.
The $Z$ component of electric field at $(0,0,Z)$ is found to be $E_Z=\frac{\sigma }{f{ϵ}_0}[1-\frac{Z}{\sqrt{R^2+Z^2}}]$
Then, the value of $2f$ is

Answer 1

Potential due to a uniformly charged disc at a perpendicular distance $Z$ from its centre is, $V=\frac{\sigma }{2{ϵ}_0}[\sqrt{R^2+Z^2}-Z]$
Here, $\sigma$ is surface charge density.

Potential due to a quarter disc is given by, $V={\frac{1}{4}}^{\text{th}}$of the potential due to complete disc with charge density $\sigma$.
$\Rightarrow V=\frac{\sigma }{8{ϵ}_0}[\sqrt{R^2+Z^2}-Z]$

The relation between Electric field and Electric potential is mathematically given by-
$\Rightarrow E_Z=-\frac{dV}{dZ}=\frac{\sigma }{8{ϵ}_0}[1-\frac{Z}{\sqrt{R^2+Z^2}}]$