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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.

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  • \$\begingroup\$ Hi! EE.SE likes \$ instead of $ for MathJax. I've corrected the formatting, but please take a look to make sure I didn't break your equations. Thanks :) \$\endgroup\$ – bitsmack Mar 29 '16 at 22:58
  • \$\begingroup\$ Actually, it looks like a few of us did it concurrently! Hope it turned out all right... \$\endgroup\$ – bitsmack Mar 29 '16 at 23:02
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I see my error. The intensity in the direction from a point charge on the circle to \$z = \pm 1 m\$ is \$4.5 V/m\$. Divide by \$\sqrt{2}\$ or multiply by \$\cos \pi/4\$ to the the component of intensity in the \$z\$ direction. \$4.5 / \sqrt{2} \approx 3.18\$.

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