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 $O$?

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.

(a)

(b)

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 $(e^{i 2\pi / N})^N =$ $e^{i 2\pi} =$ $(\cos 2\pi, \sin 2\pi) = 1$.