Question

Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.

Answer 1

Let the total charge of the ring be Q.
Radius of the ring = R
The electric field at distance x from the centre of ring,
$E=\frac{Qx}{4\pi {\epsilon }_0{\left(R^2+x^2\right)}^{3/2}}...\left(1\right)$
For maximum value of electric field,
$\frac{dE}{dx}=0$
From equation (1),
$\frac{dE}{dx}=\frac{Q}{4\pi {\epsilon }_0}\left[1(R^2+x^2{)}^{-3/2}-x{\displaystyle \frac{3}{2}}(R^2+x^2{)}^{-5/2}2x\right]=0$
$$\begin{gathered} \Rightarrow R^2+x^2-3x^2=0 \\ \Rightarrow 3x^2=R^2 \\ \Rightarrow x=\frac{R}{\sqrt{2}} \end{gathered}$$