Question

Find the electric field on the axis of a uniformly charged disc.

Answer 1

Consider a disc of uniform surface charge density $\sigma$.
Let us calculate the electric field due to a charged ring of radius $r$, from the centre and having a width, $dr.$ Here we are using the expression for electric field intensity for a charged ring of radius $r$ at a point on the axis at a distance $x$ from the centre,
$E=\frac{kxq}{(x^2+r^2{)}^{3/2}}$ directed along the axis outwards from the centre.
Thus electric field due to elementary ring,
$dE=\frac{kxdq}{(x^2+r^2{)}^{3/2}}$, directed along the line $OP$.
Now, the area of the ring, $dS=2\pi r.dr.$
Charge on the elementary ring
$⟹dq=\sigma 2\pi ede$,
Thus, electric field due to entire disc
$|E{|}_0=\int dE=kx\stackrel{R}{\underset{0}{\int }}\frac{\sigma 2\pi rdr}{(x^2+r^2{)}^{3/2}}=\frac{1}{4\pi {\in }_0}x.\sigma \pi \stackrel{R}{\underset{0}{\int }}\frac{2r}{(x^2+r^2{)}^{3/2}}dr$
$E=\frac{\sigma }{2{\in }_0}[1-\frac{x}{(x^2+R^2{)}^{1/2}}]=\frac{\sigma }{1{\epsilon }_0}(1-\cos \theta )$, where $\theta =$ semi vertical angle subtend by the disc at $R$.
Also, as $R⟶\infty ,E=\frac{\sigma }{2{\in }_0}$, which is the electric field in front of an infinite plane sheet of charge,