Question

. At what distance along the central axis of a ring of radius R and uniform charge Q is the magnitude of the electric field maximum?

Answer 1

We know that

$E=\frac{kqx}{(x^2+R^2)^\frac{3}{2}}$

$\frac{dE}{dy}=0$

$\frac{dE}{dy}=Kq\frac{(x^2+R^2)^\frac{3}{2}-\frac{3}{2}(x^2+R^2)^\frac{1}{2}.2x^2}{x^2+R^2}$

$Kq\frac{(x^2+R^2)^\frac{3}{2}-\frac{3}{2}(x^2+R^2)^\frac{1}{2}.2x^2}{x^2+R^2}=0$

$${(x^2+R^2)^\frac{3}{2}-\frac{3}{2}(x^2+R^2)^\frac{1}{2}.2x^2}=0$$

$x^2+R^2=3x^2$

$2x^2=R^2$

$x=\frac{R}{\sqrt{2}}$

When distance $x=\frac{R}{\sqrt{2}}$ Electric field is maximum