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. $E =\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{ Z }{\left( Z ^{2}+ R ^{2}\right)^{1 / 2}}\right)$
• B. $E =\frac{\sigma}{2 \varepsilon_{0}}\left(1+\frac{ Z }{\left( Z ^{2}+ R ^{2}\right)^{1 / 2}}\right)$
• C. $E =\frac{2 \varepsilon_{0}}{\sigma}\left(\frac{1}{\left( Z ^{2}+ R ^{2}\right)^{1 / 2}}+ Z \right)$
• D. $E =\frac{\sigma}{2 \varepsilon_{0}}\left(\frac{1}{\left( Z ^{2}+ R ^{2}\right)}+\frac{1}{ Z ^{2}}\right)$

Answer 1

Answer: (A)

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

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