Categories

# 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 $$z=\frac{r}{\sqrt{2}}$$

## By Dr. Ashani Dasgupta

Ph.D. in Mathematics, University of Wisconsin, Milwaukee, United States.

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

Founder, Cheenta

This site uses Akismet to reduce spam. Learn how your comment data is processed.