At what distance along the central axis of a uniformly charged plastic disc of radius R is the magnitude of the electric field equal to one-half the magnitude of the field at the centre of the surface of  the disc?

At a point on the axis of uniformly charged disc at a distance x from the centre of the disc, the magnitude of the electric field is,

$E=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{x}{\sqrt{x^2+R^2}}\right]$

At centre,   $E_c=\frac{\sigma }{2{\epsilon }_0}$

Given that,    $\frac{E}{E_c}=\frac{1}{2}$

Then,    $1-\frac{x}{\sqrt{x^2+R^2}}=\frac{1}{2}$

or    $\frac{x}{\sqrt{x^2+R^2}}=\frac{1}{2}$

On squaring both sides, we get

$x^2=\frac{x^2}{4}+\frac{R^2}{4}$

Thus,    $x^2=\frac{R^2}{3}\Rightarrow x=\frac{R}{\sqrt{3}}$