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

Answer 1

The correct option is B $E=\frac{\sigma }{2{\epsilon }_0}(1-\frac{Z}{(Z^2+R^2{)}^{1/2}})$

Consider a small ring of radius $r$ and thickness $dr$ as shown.


Charge on the elemental ring is :
$dq=\sigma dA=\sigma \times 2\pi rdr=2\pi \sigma rdr$
Electric field due to this ring at a distance $Z$ is :
$dE=\frac{dqZ}{4\pi {ϵ}_o(r^2+Z^2{)}^{3/2}}$
Therefore, electric field due to the whole disc is given by :
$E=\int dE={\int }_0^q\frac{dqZ}{4\pi {ϵ}_o(r^2+Z^2{)}^{3/2}}$

$E=\frac{2\pi \sigma Z}{4\pi {ϵ}_o}{\int }_0^R\frac{rdr}{(r^2+Z^2{)}^{3/2}}$

$\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}(1-\frac{Z}{(Z^2+R^2{)}^{1/2}})$