Q 12-01-123JEE MainJEE Main 2021 (27 Aug, Shift 1)Medium
A uniformly charged disc of radius $R$ having surface charge density $\sigma$ is placed in the $xy$ plane with its center at the origin. Find the electric field intensity along the $z$-axis at a distance $Z$ from origin:
Answer: (C) $E = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{Z}{(Z^2+R^2)^{1/2}}\right)$
Divide the disc into rings of radius $r$ and width $dr$ (charge $dq = \sigma\,2\pi r\,dr$). Each gives an axial field $\dfrac{k\,dq\,Z}{(r^2+Z^2)^{3/2}}$:
$$E = \frac{\sigma Z}{2\varepsilon_0}\int_0^R\frac{r\,dr}{(r^2+Z^2)^{3/2}} = \frac{\sigma Z}{2\varepsilon_0}\left[\frac{1}{Z} - \frac{1}{\sqrt{Z^2+R^2}}\right] = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{Z}{\sqrt{Z^2+R^2}}\right)$$
Solution by Sreeraj P, M.Sc Physics