# Force on a test charge placed at the center of a 13-sided regular polygon?

Assume a test charge with charge $Q$ is placed at the center of a 13-sided regular polygon, the vertices of which are charges with equal charge $q$. What is the electrical force on the center charge?

To start we write this as our electrical force... $$\vec{F}_{Q} = \sum^{13}_{n} \frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{R^{2}} \left(cos\left(\frac{2n\pi}{13}\right)\hat{i} , sin\left(\frac{2n\pi}{13}\right)\hat{j}\right)$$

and with some simplification we get... $$\vec{F}_{Q} = \frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{R^{2}} \left\{1-2cos\left(\frac{\pi}{13}\right) + 2cos\left(\frac{2\pi}{13}\right)-2cos\left(\frac{3\pi}{13}\right) + 2sin\left(\frac{\pi}{26}\right)-2sin\left(\frac{3\pi}{26}\right)+2sin\left(\frac{5\pi}{26}\right), 0\right\}$$ ...more simplification... $$\vec{F}_{Q} = \frac{1}{4\pi\varepsilon_{0}}\frac{qQ}{R^{2}} \left\{ 1.11022\times10^{-16}, 0 \right\}$$ Now we have an electrical force on the test charge, however, in a ring of equidistant charges such as the one mentioned above, shouldn't the $\hat{i}$ component be equal to zero? Does anyone have any ideas as to why I'm getting a non-zero number for my $\hat{i}$ component?

• A sprinkling of backslashes on your \coses and \sines wouldn't come amiss. – Emilio Pisanty Mar 4 '14 at 23:54

$\def\l{\left}\def\r{\right}$ You can work in the complex plane. The real component is your $\hat i$-component the imaginary component is your $\hat j$-component. There $\l(\cos(\phi),\sin(\phi)\r)$ is represented as $\exp(i\phi)$ with $i=\sqrt{-1}$.

In this context your sum is $$\sum_{k=0}^{N-1} e^{i2\pi\frac kN}.$$ if we substitute $a := e^{i\frac{2\pi}N}$ into this formula we obtain $$\sum_{k=0}^{N-1} a^k$$ which can be transformed into a telescope sum by multiplying with $(1-a)$ $$(1-a)\sum_{k=0}^{N-1} a^k = \sum_{k=0}^{N-1} a^k - \sum_{k=1}^{N} a^k = 1-a^N$$ Thus, we have $$\sum_{k=0}^{N-1} a^k = \frac{1-a^N}{1-a}$$ But $1-a^N = (e^{i\frac{2\pi}N})^N = e^{i2\pi} = 1$ and we obtain $$\sum_{k=0}^{N-1} e^{i2\pi\frac kN} = 0.$$

Your sum is zero in exact arithmetics.

You simplified incorrectly, the result is zero. I assume you computed it using floating point arithmetic, which explains why the discrepancy is on the order of machine epsilon. Try using symbolic manipulation in Mathematica:

Sum[Cos[2 \[Pi] n/13], {n, 13}] // FullSimplify

Out: 0


Also you can use the Langrange identity

$$\sum_{n=1}^{N} \cos (n \theta) = - \frac{1}{2} + \frac{\sin \left( N + \frac{1}{2} \right) \theta}{ 2 \sin \left( \frac{\theta}{2}\right)}$$

which yields

$$\sum_{n=1}^{13} \cos \left( \frac{2 n \pi}{13} \right) = -\frac{1}{2}+\frac{\sin \left( \left(13+\frac{1}{2}\right) \frac{2 \pi }{13}\right)}{2 \sin \left(\frac{\pi }{13}\right)} = -\frac{1}{2} +\frac{\sin( 2 \pi) \cos \left( \frac{\pi}{13} \right) + \sin \left( \frac{\pi}{13} \right) \cos( 2 \pi)}{2 \sin \left(\frac{\pi }{13}\right)} =0$$

I find this more appealing than saying the sum must be zero due to symmetry

I think we have here simple analytical geometry valid for any n-sided regular polygon : The sides as vectors $$\;\mathbf{s}_k\;$$ sum up to zero $$\;\sum\mathbf{s}_k=\boldsymbol{0}$$. The radii as vectors $$\;\mathbf{r}_k\;$$ sum up to zero too $$\;\sum\mathbf{r}_k=\boldsymbol{0}$$. The radii vectors form a n-sided regular polygon similar to the initial one. In case of a regular hexagon the so formed hexagon is identical to the initial one. In case of a regular decagon the similarity ratio is equal to the golden ratio $$\;\phi=\left(1+\sqrt{5}\right)/2$$.

A non-zero resultant force would appear under a rotational symmetry breaking (for example by removing a charge $$\;q\;$$ from one vertex).