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Oct 6, 2017 at 13:51 comment added Alexandre Eremenko @KlausDraeger: This is what I referred to as "general considerations" in my answer: arguments which use very little of the nature of the potential.
Oct 6, 2017 at 13:00 comment added Klaus Draeger @JoshuaBenabou How about the following? If you consider a sufficiently large ball $B$ which contains all the points in its interior, it is clear that the gravitational field $F$ is continuous on $B$ and pointing into $B$ everywhere on its surface. Then for a sufficiently small $\varepsilon$, $x+\varepsilon F(x)$ is in $B$ for all $x\in B$. As an endomorphism of a convex compact subset, it has a fixpoint (where $F$ has to be $0$) by the Schauder fixpoint theorem.
Oct 6, 2017 at 8:51 comment added math_lover @AlexandreEremenko: Why is it obvious that the number of points is nonzero? For example even with $N=3$, if the configuration does not consist of 3 collinear points or an equilateral triangle, the existence of a point $P$ where the field vanishes is not obvious.
Oct 6, 2017 at 3:09 comment added Steven Stadnicki @KonstantinosKanakoglou At its root, the question is just about Newtonian $r^{-2}$ potential fields; aside from charge, there's no real difference between gravity and EM fields in that regard.
Oct 6, 2017 at 3:08 comment added zibadawa timmy @KonstantinosKanakoglou Depends on your definition of positive and the scalar potential.
Oct 6, 2017 at 2:56 history edited Alexandre Eremenko CC BY-SA 3.0
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Oct 5, 2017 at 23:32 comment added Konstantinos Kanakoglou "Since you are talking about gravitation (rather than electrostatics) I assume that all charges are positive": isn't this statement a little shady? Given that gravitational forces are always attractive?
Oct 5, 2017 at 23:01 history edited Alexandre Eremenko CC BY-SA 3.0
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Oct 5, 2017 at 22:54 history answered Alexandre Eremenko CC BY-SA 3.0