Electric field at the center of a polygon: symmetry!

Posted by Diego Assencio on 2013.11.26 under Physics (Electromagnetism)

Consider a regular polygon with $N$ sides (and therefore $N$ vertices) such that the distance between its center $O$ and each of its vertices is $r$. At each vertex of the polygon, a charge $q$ is placed (see figure 1). What is the electric field ${\bf E}$ produced by these charges at the point $O$?

Fig. 1: Regular pentagon with a charge $q$ placed on each of its vertices. In this case, $\theta = 2\pi / 5$. For a general regular polygon with $N$ sides, $\theta = 2\pi / N$.

If your strategy to solve a problem of this kind is to actually compute ${\bf E}$ by hand, you missed an important fact: there is symmetry in this problem. Symmetry is a very important ingredient in physics: it can make many complicated-looking problems become easily solvable in a few lines (or, in our case, in one line).

To see what I mean by that, imagine the electric field produced by the charges at the point $O$ is as shown in figure 2a.

Charges placed on vertices of a regular polygon
Charges placed on vertices of a regular polygon
Fig. 2: Rotating a pentagon by $\theta = 2\pi/5$ around its center point in the anti-clockwise (or clockwise) direction produces the same charge configuration (b) as we initially had in (a). In general, this is true if one rotates a polygon with $N$ sides by $\theta = 2\pi/N$ around its center point. Since the final charge configuration is the same as the initial one, the electric field ${\bf E}$ at the point $O$ must be the same before and after the rotation (and therefore it must have zero magnitude at that point).

If we now rotate the polygon by $\theta = 2\pi/N$ clockwise or counterclockwise we will obtain exactly the same charge configuration as we had before (compare figures 2a and 2b). In other words, the electric field ${\bf E}$ at the point $O$ must be the same before and after the rotation is applied. But if the whole polygon is rotated, ${\bf E}$ must also be rotated with it. We will have an absurd situation unless ${\bf E} = {\bf 0}$. In other words, the electric field at $O$ must be zero!

A reader who prefers things "proven mathematically" might be bothered with this type of proof. As a matter of fact, some might not even consider it a proof at all. However, arguments of this kind are ubiquitous in Phyics and yield correct results with very little work.

In any case, if you are one of those readers I just mentioned, let's compute ${\bf E}$ by hand to bring peace back to your mind. Numbering each charge in a counterclockwise manner (as in figure 1), the $k$-th charge would then be located at: $$ {\bf x}_k = r(\cos\theta_k, \sin\theta_k) = r(\cos(2\pi k/N), \sin(2\pi k/N)) $$ for $k = 1, 2, \ldots, N$. Since $e^{i\theta} = (\cos\theta,\sin\theta)$ (a complex number is a point in $\mathbb{R}^2$), then we can write: $$ {\bf x}_k = r e^{i 2\pi k / N} $$ This will allow us to compute ${\bf E}$ more easily. If ${\bf E}_k$ is the electric field produced by the $k$-th charge at the point $O$, then: $$ {\bf E}_k = \displaystyle\frac{kq}{\|{\bf x}_k\|^3}(-{\bf x}_k) = -\displaystyle\frac{kq}{r^3} r e^{i 2\pi k / N} = -\displaystyle\frac{kq}{r^2} e^{i 2\pi k / N} $$ and therefore: $$ {\bf E} = \sum_{k=1}^{N} {\bf E}_k = \sum_{k=1}^N \left( -\displaystyle\frac{kq}{r^2} e^{i 2\pi k / N} \right) = -\displaystyle\frac{kq}{r^2} \sum_{k=1}^N (e^{i 2\pi / N})^k \label{post_1d16b3e5666a13fc6cfa572b64f47dd2_E_as_sum} $$ But since for any complex number $z \neq 1$ we have: $$ \sum_{k=0}^N z^k = \displaystyle \frac{1 - z^{N+1}}{1 - z} \Longrightarrow \sum_{k=1}^N z^k = \displaystyle \frac{1 - z^{N+1}}{1 - z} - 1 = \displaystyle \frac{z - z^{N+1}}{1 - z} $$ then, from equation \eqref{post_1d16b3e5666a13fc6cfa572b64f47dd2_E_as_sum} with $z = e^{i 2\pi/N}$ we obtain: $$ {\bf E} = -\displaystyle\frac{kq}{r^2} \displaystyle\frac{e^{i 2\pi / N} - (e^{i 2\pi / N})^{N+1}}{1 - e^{i 2\pi / N}} = -\displaystyle\frac{kq}{r^2}e^{i 2\pi / N}\frac{1 - (e^{i 2\pi/N})^N}{1 - e^{i 2\pi / N}} = {\bf 0} $$ since $(e^{i 2\pi / N})^N =$ $e^{i 2\pi} =$ $(\cos 2\pi, \sin 2\pi) = 1$.


Shine on Mar 29, 2017:
Katyaini on May 14, 2017:
Okkkay... What happens if instead of discrete charges, a charge is distributed over the wire forming the Pentagon.???
Diego Assencio on May 14, 2017:
@Katyaini: The symmetry argument still applies: if you rotate the pentagon by $\theta = 2\pi / 5$ (clockwise or anti-clockwise), you will end up with the same charge configuration you had initially. If the electric field ${\bf E}$ at the center of the pentagon were not zero, it would rotate by $\theta = 2\pi / 5$ as well and end up pointing along a different direction as it initially did, which is impossible since the charge configurations before and after the rotation are the same. Therefore, ${\bf E} = {\bf 0}$ in this case as well.

Another way to see this system is as a collection of pentagons with point charges on their vertices: each pentagon will individually produce a zero electric field at the center, and therefore the total electric field is zero.
pallab on Jun 10, 2017:
what happens to the electric field at center of fig.1 if one charge say 1(vertex)is omitted?
Diego Assencio on Jun 11, 2017:
@pallab: Removing a charge $q$ is equivalent to adding a charge $-q$ on top of it. In both situations, the total charge at that point becomes zero.

Therefore, the electric field resulting from removing a charge is the same as the one produced by all charges originally placed at the vertices of the polygon, plus the one produced by the added charge $-q$. Since the original charges produce no electric field at the center of the polygon, the total electric field at that point is then simply the one produced by $-q$.
Steven Stam on Feb 03, 2018:
I'm working on a similar problem where after proving E is 0 at the center of a regular polygon, you also have to find all other enclosed points where E=0. I'm trying to do this by writing a generalized formula for E from a given charge and then taking the vector sum and setting it equal to 0. This is a lot of work but I can't find a way to use symmetry about a point which is not the center. Anyone have any ideas on reducing the calculations?
Mohit Singh on Sep 23, 2018:
Does this symmetry argument applies to points other than centre of the polygon , will field be zero at any point inside the polygon?
Diego Assencio on Sep 24, 2018:
@Mohit: ${\bf E}$ cannot be zero outside the polygon. To see that, choose any point outside the polygon, then draw a line passing through that point and the center of the polygon. Assuming $q \gt 0$ (the argument is similar for $q \lt 0$), consider the electric field ${\bf E}_k$ produced by the $k$-th charge at the point you chose: it will always have a component along the line which points away from the center of the polygon. The sum of all such components can therefore not be zero, so ${\bf E}$ will have a nonzero radial component there. With a similar argument, you can convince yourself that ${\bf E} \neq {\bf 0}$ on the surface of the polygon as well.

Inside the polygon, things become trickier. You could compute each ${\bf E}_k$ explicitly and then add them together, but I think the final expression will not be easy to analyze. One suggestion is to plot the magnitude of ${\bf E}$ and see what it looks like.

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