Question

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 :-

• A. $E=\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{Z}{{\left(Z^2+R^2\right)}^{1/2}}\right)$
• B. $E=\frac{\sigma }{2{\epsilon }_0}\left(1+\frac{Z}{{\left(Z^2+R^2\right)}^{1/2}}\right)$
• C. $E=\frac{2{\epsilon }_0}{\sigma }\left(\frac{1}{{\left(Z^2+R^2\right)}^{1/2}}+Z\right)$
• D. $E=\frac{\sigma }{2{\epsilon }_0}\left(\frac{1}{\left(Z^2+R^2\right)}+\frac{1}{Z^2}\right)$

Answer 1

Answer: (A)

Consider a small ring of radius $r$ and thickness $dr$ on disc.

area of elemental ring on disc
$dA=2\pi rdr$
charge on this ring $dq=\sigma dA$
$dEz=\frac{kdqz}{{\left(z^2+r^2\right)}^{3/2}}$
$E=\underset{0}{\overset{R}{\int }}d{E}_z=\frac{\sigma }{2{\in }_0}\left[1-\frac{z}{\sqrt{R^2+z^2}}\right]$