On the existence of a point in the plane where repulsive central forces exerted by $ n $ fixed points cancel

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This is a physics-inspired question.

In what follows, $ \alpha \in (1,\infty) $ is a fixed constant, $ n \in \mathbb{N} $ a fixed integer $ \geq 2 $, and $ [n] \stackrel{\text{df}}{=} \mathbb{N}_{\leq n} $.


Let $ P_{1},\ldots,P_{n} $ be distinct points in the plane $ \mathbb{R}^{2} $. Let $ q_{1},\ldots,q_{n} $ be positive real numbers. Then for every $ i \in [n] $, define a vector-valued function $ \mathbf{F}_{i}: \mathbb{R}^{2} \setminus \{ P_{i} \}_{i \in [n]} \to \mathbb{R}^{2} $ by $$ \forall X \in \mathbb{R}^{2} \setminus \{ P_{i} \}_{i \in [n]}: \quad {\mathbf{F}_{i}}(X) \stackrel{\text{df}}{=} - \frac{q_{i}}{\left\| \overrightarrow{X P_{i}} \right\|^{\alpha}} \cdot \overrightarrow{X P_{i}}. $$ (If $ \alpha = 3 $, then we can view $ q_{i} $ as a positive electrical charge carried by $ P_{i} $ and interpret $ {\mathbf{F}_{i}}(X) $ as a repulsive electrostatic force exerted on $ X $ by $ P_{i} $. Even if $ \alpha \in (1,3) \cup (3,\infty) $, we can still interpret $ {\mathbf{F}_{i}}(X) $ as some sort of repulsive central force exerted on $ X $ by $ P_{i} $.)

Question. Does there exist an $ X \in \mathbb{R}^{2} \setminus \{ P_{i} \}_{i \in [n]} $ such that $ \displaystyle \sum_{i \in [n]} {\mathbf{F}_{i}}(X) = \mathbf{0} $? In more physical terms, is there a point $ X $ at which the repulsive forces cancel? A rigorous argument is desired.

One thing is for sure. If such an $ X $ exists, then it must lie within the closed convex hull of $ \{ P_{i} \}_{i \in [n]} $.

Thank you for your help!

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I think there must be stationary points. Consider a closed curve surrounding the vector field far from the sources of the vector field. The Index of the vector field around this curve must be $1$ (the vector field rotates one time counterclockwise along the curve). By Poincarè-Hopf theorem the indexes of the singular points inside the curve must sum up to +1 (the index along the curve) and since you have $n$ sources with indexes that sum up to a total of $+n$ to get $+1$ we need to have a $-n+1$ contribution i.e. at least another critical saddle point.

An online reference could be this one and also this from "Visual Complex Analysis" (starting from pag. 459).