Points where electric field is zero when charges are present at vertices of a regular polygon There is a $n$-sided regular polygon with a charge $q$ at each vertex. I know that there are $n$ points, other the center of the polygon, where the electric field is zero. But why is this so? Is there a general way to prove it?
PS: I know some questions related to my question have been asked, but none of them gives me a satisfactory reason why there should be a total of $n+1$ neutral points in space for such a charge distribution.
 A: Consider a line connecting two adjacent charges on your polygon, and then another line which bisects that one, going in the positive x direction through the center of the polygon (at x = 0). To show that your premise is true, you would need to find one (and only one, if (n) is odd) other point (other than at x = 0) on that line of symmetry, where the sum of the x components of the fields is zero. For a general proof, the distances and components of distances would be a function of (n), and would depend on whether (n) was odd or even.
A: *

*For what it's worth, 2D Morse theory (with the assumption that all critical points for the electric potential are non-degenerate) yields that
$$c_1-c_0~=~n-1\qquad\text{and}\qquad c_2~=~0,$$
where
$$\begin{align} c_0~:=~& \#{\rm minimum~pts}, \cr 
c_1~:=~& \#{\rm saddle~pts},  \cr 
c_2~:=~& \#{\rm maximum~pts},\end{align}$$
cf. e.g. my Phys.SE answer here.


*So under these assumptions$^1$ OP's claim for the regular $n$-polygon$^2$
$$\begin{align}V(z)~=~&\sum_{j=1}^n \left|z-\exp\frac{2\pi i j}{n}\right|^{-1}\cr
~=~&\sum_{j=1}^n \left(\left(x-\cos\frac{2\pi  j}{n}\right)^2+\left(y-\sin\frac{2\pi  j}{n}\right)^2\right)^{-1/2}\cr
~=~&\sum_{j=1}^n\left(1+r^2-2r\cos\left(\theta-\frac{2\pi j}{n}\right)\right)^{-1/2}
\end{align} $$
that the number of critical points are $c_2+c_1+c_0=n+1$ would follow if we can show that $c_0=1$, i.e. that the center is the only local minimum.

$^1$ It is not difficult to see that the center $r=0$ is a non-degenerate critical point:
$$ V(r,\theta)~=~\left\{ 
\begin{array}{rl}
n\left(1+\frac{r^2}{4}\right) +{\cal O}(r^3) &\text{minimum pt for } n\geq 3,\cr 2+\frac{r^2}{2}\left(1+3\cos 2\theta\right)+{\cal O}(r^3)&\text{saddle pt for } n=2. \end{array}\right. $$
$^2$ We assume that $n\geq 3$. If $n=1$, then $c_0=0=c_1$. If $n=2$, then $c_0=0$ and $c_1=1$.
