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[INTRO]

In such arrangement of identical charged particles named $P_1$, $P_2$, ..., $P_n$ (In this diagram $n=5$), the electric field in the center is always zero.

arrangement

The electric field is given by this equation:

$$\vec F=k\frac{q}{r^2}\hat r$$

Where $\hat r$ points to the radial direction.

If $2|n$, then the forces are obvious canceled by symetry.

If $n$ is odd, again the components of forces in the $y$ axis cancel by symetry. But for the $x$ direction the components cancel if this condition is true:

$$\sum_{c=0}^{n-1}\cos(c \theta)=0 \ \ \ \ \ \ \theta=\frac{2\pi}n$$

[END OF INTRO]

My main problem is to prove for any $n$, this relation holds:

$$\sum_{c=0}^{n-1}\cos(2\pi \frac{c}{n})=0$$

AHB
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  • Hint: the real part of a finite geometric series. – GEdgar Aug 21 '16 at 16:46
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    Write it in terms of complex exponential and use the finite geometric series formula. – Ian Aug 21 '16 at 16:46
  • @Ian $e^{\theta i}$ you mean? I am not much familiar with complex world. – AHB Aug 21 '16 at 16:48
  • Intuitively, if all of those radii are vectors, they cancel out and equal the zero vector. That means that, not only do the vertical components cancel (which is easy to see independently, because $\sin(-x)=-\sin(x)$), but so do the horizontal components. – G Tony Jacobs Aug 21 '16 at 16:53
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    @GTonyJacobs Of course! But one will have to merge that intuition with a proof if he wants to make sure that the intuition is right. – AHB Aug 21 '16 at 16:56
  • I wasn't arguing against proof-writing, @AHB. Another perspective is that of symmetry: If there is a net pull in any direction, then by rotating our coordinates repeatedly through an angle of $\theta$, that net pull should be in each of 5 different directions! – G Tony Jacobs Aug 21 '16 at 16:58
  • @JMoravitz I think this question is more likely to be in the search queue than that one is. It also includes geometrical and physical uses. – AHB Aug 21 '16 at 16:58
  • @AHB That one is already half a year old, has a nearly a dozen upvoted answers, and a large number of views already and successfully answers your underlying question. If you think it does not fully answer your question, then please specify why not so that we may fill that gap in understanding. – JMoravitz Aug 21 '16 at 17:00
  • @JMoravitz hmmm. Yeah your right. I guessed such a question must have been asked before. I just couldn't find the keyword to search. Then mark as duplicate or delete? – AHB Aug 21 '16 at 17:03
  • @AHB It is certainly difficult sometimes to search for and find duplicates from some time ago. That is one of the benefits from being active for so long on the site, we can sometimes remember duplicates manually (as was the case for me here, I know I answered this in the past and could check my own history). If you are confident that your phrasing of the question could be more frequently searched, then it may indeed be helpful to leave it up as duplicate for other users to find the linked question. Otherwise, if your only goal was the answer deletion is fine too. I leave that up to you. – JMoravitz Aug 21 '16 at 17:07
  • Thanks @JMoravitz for having found this very pertinent reference. – Jean Marie Aug 21 '16 at 17:08
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    The force $F_1$ at the orgin exerted by $P_1$ is horizontal and makes an angle of $\pi$ with respect to the positive axis. Add to that the force $F_2$ exerted by $P_2,$ which has the same magnitude but makes an angle of $\pi+2\pi/5$ with respect to the positive axis. Keep going. Adding $F_1 + \cdots +F_5$ as vectors of course means starting with $F_1$ pointing from at $0,$ starting $F_2$ at the end of $F_1,$ etc. What you'll see is that the vector sum leads you around the perimeter of a regular pentagon. We started at $0,$ and we have to end at $0!$ – zhw. Aug 21 '16 at 17:46
  • Do you agree that the total force at the center should be independent of the orientation of this arrangement of charges? If so, then it must be the case that the $x$-components cancel since you can swap the roles of the $x$- and $y$- axes without changing the total charge. – amd Aug 21 '16 at 23:12
  • @amd That is an easily understood non-mathematical proof. Proof by contradiction. Thanks. – AHB Aug 22 '16 at 08:47

1 Answers1

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This is valid for any regular polygon. It can be viewed physically:

Sum of cosines or sines of equi-spaced forces is zero by finding resultant using a statics force polygon vector diagram.

Or mathematically:

Formula for sum of $n$ equi-spaced ( interval $\alpha , \alpha = 2 \pi/n $) sines. It can be derived by summing terms from De Moivre's theorem as a geometric progression. $$ \frac { \sin n \alpha/2}{ \sin \alpha/2}\cdot \sin ( average\, angle) $$

Sum of $n$ equi-spaced ( interval $\alpha $) cosines

$$ \frac { \sin n \alpha/2}{ \sin \alpha/2}\cdot \cos ( average\, angle) $$

Quantity in numerator is $ \sin n \pi $ that vanishes

Narasimham
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