Question

At what distance along the central axis of uniformly charged plastic disk of radius R is magnitude of electric field equal to half the electric field at the centre of the disk

Answer 1

Suppose,at distance x along the central axis is the magnitude of the electric field is one half the magnitude of the field at the center of the surface of the disc.
Then,
E(x)=1/2 E(0)
$$\begin{gathered} \frac{\sigma }{2{\in }_0}[1-\frac{x}{\sqrt{x^2+R^2}}]=\frac{\sigma }{4{\in }_0} \\ or \\ [1-\frac{x}{\sqrt{x^2+R^2}}=\frac{1}{2} \\ or \\ \frac{x}{\sqrt{x^2+R^2}}=\frac{1}{2} \\ or2x=\sqrt{x^2+R^2} \\ Onsquaringbothsides, \\ 4x2=x2+R23x2=R2x=R/3 \end{gathered}$$