Question

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

Answer 1

The electric field at distance x from the centre of ring is
$E=\frac{Q}{4\pi {ϵ}_0(R^2+x^2{)}^{\frac{3}{2}}}$
For maximum value of eelctric field
$\frac{dE}{dx}=0$
From equation (i)
$\frac{dE}{dx}=\frac{Q}{4\pi {ϵ}_0}$
$\left[1.\right(R^2+x^2{)}^{\frac{3}{2}}-x.\frac{3}{2}(R^2+x^2{)}^{\frac{1}{2}}.2x]=0or{R}^2+x^2-3x^2=0or2x^2=R^2\therefore x=\frac{R}{\sqrt{2}}$