# $2n$ point charges on a circle at $O$, what's total electric field intensity $\overline{E}$ at $\pm 1\mathrm{m}$?

The problem reads is from a Schaum's Outlines of EMFT and reads:

A total charge of $1\mathrm{nC}$ is equally distributed among $2n$ points which are placed equidistantly on a circle with $1\mathrm{m}$ radius centered at the origin in the $xy$ plane. Find the electric field intensity on the axis of the circle at $z = \pm 1\mathrm{m}$.

The answer given is $\overline{E} = 3.18 \hat{a}_z \ \mathrm{Vm^{-1}}$. But I keep getting $\overline{E} = 4.5 \hat{a}_z \ \mathrm{Vm^{-1}}$.

My method: By symmetry of a $2n$-gon at the origin, the field intensities parallel to the $xy$ plane will cancel each other out when you add them all together, so that the intensity experienced by something at $z = \pm 1\mathrm{m}$ is all due to intensity along $\hat{a}_z$.

So far we have:

$$\overline{E} = 2n \times \dfrac{\dfrac{10^{-9}}{2n}}{4\pi\dfrac{10^{-9}}{36 \pi} (\sqrt{2})^2} \hat{a}_z = \pm 4.5 \hat{a}_z \ \mathrm{Vm^{-1}}$$

where $\epsilon \approx 10^{-9}{36 \pi}$ is the permittivity.

I think it may be because I'm using $\sqrt{2}$ for distance when I should be using $1$? But that would give $9 \ \mathrm{Vm^{-1}}$, so I'm not sure what to do. Thanks.