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# Maximum Electric Field of a Ring

A ring of radius $$r$$ is located in the $$x-y$$ plane is given a total charge $$Q=2\pi R\lambda$$. Show that $$E$$ is maximum when the distance $$z=r/\sqrt{2}$$.

Discussion:

The elctric field at distance z from the centre of the ring on the axis of the ring with charge $$Q=2\pi R\lambda$$ is given by $$E=\frac{\lambda r}{2\epsilon_0}\frac{z}{(z^2+r^2)^{3/2}}$$

The maximum field is obtained by setting $$\frac{dE}{dz}=0$$

This gives $$(z^2+r^2)^{1/2}(r^2-2z^2)=0$$
Since the first factor cannot be zero for any real value of z, the second factor gives $$z=\frac{r}{\sqrt{2}}$$

September 24, 2017