Background:
During my physics class we were talking about electric fields. The field around a point particle with charge $q$ is $\left|\frac{Kq}{r^2}\right|$ where $r$ is the distance from the particle and $K$ is a constant (so can be discarded for this matter), and the direction of the field is away from the particle (or towards if it's negatively charged). Electric fields are addative, so for a system with multiple charges you just add all their individial fields.
One of the practice problems we solved involved three charges $+2q, -q, -q$ arranged in a triangle, and we were able to find a point where all the force vectors cancel each other out, and the field there is $0$. It seems like this happens for any arrangement of the charges, although the calculation becomes slighlty more compicated. That made me wonder about the general case:
Question:
For $n\geq2$ partices with charges $q_1,\cdots,q_n$ arranged on the plane, is there a point somewhere on the plane where their total electric field is zero? Is there a way of calculating that point? How about dimension higher than 2?
For the case of $n=2$ it's pretty easy to see I think. With $n=3$ less so, but if it works for two dimensions it'd work for any.
I could express this function as
$$E(x,y) = \sum_{i=1}^{n}{\frac{q_i}{\sqrt{(x_i-x)^2 + (y_i-y)^2}}}$$
but that only takes care of the magnitudes, I have no idea how incorporate the directions to each charge.
I realized that when "going off to infinity" the field would approach zero, which means that somewhere in the middle there's a maximum (it's a continuous function), but it's not quite what I was looking for. I'd appreciate some help starting out or finding a direction for this proof.
Best Answer
No, it's not true that there is always such a point. You need some special assumptions on the charges and their configurations. For example, take a point charge $q$ and place it at a point $\xi_0\in\Bbb{R}^n\setminus\{0\}$ and take a point charge $-q$ and place it at the diametrically opposite point $-\xi_0$. The resulting electric field is nowhere vanishing.
Next, you say
No, that's false. Yes, the electric field is continuous wherever it is defined, but the electric field is not defined at the location of the point charges themselves. The theorem you seem to be thinking of in your quote is the "extreme value theorem", which states "every continuous function on a compact set attains a maximum (and a minimum)". If you could carefully apply this to the electric potential $\phi$, you can deduce it has a local extremum somewhere, and hence its gradient at that point (which up to sign is the electric field at that point) must vanish. However, the electric potential is not defined at the positions of the point charges, so this line of reasoning fails. Once again, I refer you to the simple dipole counterexample.