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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 \(z=r/\sqrt{2}\).


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

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