1
$\begingroup$

The electric field at a point a distance $z$ above the midpoint of a segment of length $2L$ and uniform charge density $\lambda$ is given by $$\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{z\sqrt{z^2 + L^2}}\hat{\mathbf{z}} \tag{1}$$


Square loop. This is an elementary problem in Griffith's Introduction to Electrodynamics. Suppose we have four of these segments forming a square, and we want to calculate the electric field at a point $P$ a distance $z$ above the center of the square. Setting up Cartesian coordinates with origin at its center and axes parallel to its sides, considering only the side $AB$ that is parallel to the $y$ axis and lies in the first and fourth quadrant, we use $(1)$ to conclude that $$\mathbf{E}_{AB} = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{\sqrt{(z^2 + L^2)(z^2 + 2L^2)}}\hat{\mathbf{k}} \tag{2}$$ (just substituting for the new distance $z \to \sqrt{z^2 + L^2}$), where $\hat{\mathbf{k}} = \mathbf R / R$, if $\mathbf R$ is the separation vector between the point $P$ and the the midpoint of $AB$. Since $R = \sqrt{z^2 + L^2}$ and $\mathbf R = - L\hat{\mathbf x} + z\hat{\mathbf z}$, we see that $$\begin{split} \mathbf E_{AB} &= \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{\sqrt{(z^2 + L^2)(z^2 + 2L^2)}}\left(- \frac{L}{\sqrt{z^2 + L^2}}\hat{\mathbf x} + \frac{z}{\sqrt{z^2+L^2}}\hat{\mathbf z}\right) \\ &= \frac{1}{4\pi\varepsilon_0} \frac{2\lambda L}{(z^2 + L^2)\sqrt{z^2 + 2L^2}} (- L\hat{\mathbf x} + z\hat{\mathbf z}) \end{split}$$ The electric field produced by the side $CD$ has equal $z$ component and opposite $x$ component, so that we have $$\mathbf E_{AB\cup CD} = \frac{1}{4\pi\varepsilon_0} \frac{4\lambda L z}{(z^2 + L^2)\sqrt{z^2 + 2L^2}} \hat{\mathbf z} $$ Similar reasoning leads us to determining that the electric field contributed by sides $BC$ and $DA$ together is found by the same formula; therefore we conclude that, in total, $$\mathbf E = \frac{1}{4\pi\varepsilon_0} \frac{8\lambda L z}{(z^2 + L^2)\sqrt{z^2 + 2L^2}} \hat{\mathbf z}$$ When $z \gg L$ we see that the field goes like $1/z^2$ as we expected. It is also directly proportional to $8\lambda L$ which is the total charge on the loop.


Polygonal loop. Now suppose that instead the loop is shaped like an $n$-sided polygon. Let one side $AB$ be again parallel to the $y$ axis and lying in the first and fourth quadrant. If $a$ is the apothem of our polygon, formula $(2)$ becomes $$\mathbf{E}_{AB} = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{\sqrt{(z^2 + a^2)(z^2 + a^2 + L^2)}}\hat{\mathbf{k}} = \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{(z^2 + a^2)\sqrt{z^2 + a^2 + L^2}}\hat{\mathbf{k}}(- a\hat{\mathbf x} + z \hat{\mathbf z}) $$ since $\mathbf R = - a\hat{\mathbf x} + z \hat{\mathbf z}$. By a symmetry argument similar to the one I provided above, the total electric field should have no lateral component, and my conjecture is that it should be given by $$\mathbf E = \frac{1}{4\pi\varepsilon_0} \frac{2n\lambda L z}{(z^2 + a^2)\sqrt{z^2 + a^2 + L^2}} \hat{\mathbf z}$$ although I don't know the exact steps to show this formula in the general $n$ case (it should be trivial). Now, basic geometry informs us that $$a = L \cot\left(\pi/n\right)$$ so that we arrive at $$ \mathbf E = \frac{1}{4\pi\varepsilon_0} \frac{2n\lambda L z}{(z^2 + L^2\cot^2(\pi/n))\sqrt{z^2 + L^2\csc^2(\pi/n)}} \hat{\mathbf z}$$ The limit as $n \to \infty$ should give us the electric field above a circular loop of radius $L$, but Wolfram tells me it vanishes. Where is my mistake? Is my reasoning even correct in the first place?

$\endgroup$
2
  • $\begingroup$ This is really a physics question. It might be better off at the Physics Stack Exchange site? $\endgroup$
    – copper.hat
    Jan 4, 2018 at 4:57
  • $\begingroup$ One problem I foresee is that while symmetry keeps the field from having a lateral component, the analysis for odd $n$ on symmetry will require more work as the segment opposite a straight side will indeed be a corner between two angled sides instead ( a regular pentagon for example, or even an equilateral triangle), this might adjust your formula enough. But it's just a guess at this point $\endgroup$
    – Triatticus
    Jan 4, 2018 at 5:35

1 Answer 1

3
$\begingroup$

Your result gives the electric field for a polygon whose $n$ sides have length $2L$ and has then a radius $R=L/\sin(\pi/n)\to\infty$, no surprise then if $E\to0$. You should instead keep $R$ constant and plug $L=R\sin(\pi/n)$ into your formula for a polygon with $n$ sides: $$ E = \frac{1}{4\pi\varepsilon_0} \frac{2\lambda Rz\ n\sin(\pi/n) } {\big(z^2 + R^2\cos^2(\pi/n)\big)\sqrt{z^2 + R^2}}. $$ As $n\to\infty$ we have: $n\sin(\pi/n)\to\pi$ and $\cos(\pi/n)\to1$, giving thus the correct formula.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .