**EDIT:**



> 
> Given a system of $N\geq 3$ charged point particles in $\mathbb{R}^3$ of the same charge which interact according to Coulomb law (thus they repell one from each other). Is it possible that the system remains in a fixed ball all the time? (For $N=2$ this is impossible, and this is what I expect in general.)
> 
> 
> 


More precisely, denote $m\_1,\dots, m\_N>0$ the masses of the particles. Assume that the $i$th particle acts on $j$th one with the force 


$$\vec F\_{ij}=\frac{k\_ee\_ie\_j}{|\vec x\_j-\vec x\_i|^3}\cdot (\vec x\_j-\vec x\_i), $$
where $k\_e>0$ is a constant, $e\_i$ is a charge of $i$th particle such that $e\_ie\_j>0$, $\vec x\_i$ is the location of the $i$th particle. The equations of motions are
$$m\_j\frac{d^2 x\_j}{dt^2}=\sum\_{i\ne j}\vec F\_{ij}, \mbox{ where } j=1,\dots,N.\,\,\,(1)$$



> 
> The question is whether there is a solution such that for some $R$ one has
>  $$||\vec x\_i(t)||<R \mbox{ for all } t>0, \, i=1,\dots, N.$$
> 
> 
> 


**ADDED:** I expect that this is impossible. In fact I expect that not only for Coulomb law, but still in greater generality. Assume that the equations (1) are satisfied when the force $\vec F\_{ij}=\vec F\_{ij}(x\_i,x\_j)$ has the same direction as the vector $\vec x\_j-\vec x\_i$. Assume moreover that if all points are in a fixed ball of the radius $R$ then for some constant $\varepsilon >0$ such that 
$$||\vec F\_{ij}||>\varepsilon.$$
Is there a solution of (1) such that all the point are in the ball of radius $R$ for all $t>0$?



Movement of repelled particles in a ball
If all the particles remained in a bounded domain, the [virial theorem](https://en.m.wikipedia.org/wiki/Virial_theorem) would apply. In the case of a radial inverse square power law, it states that twice the asymptotic time average of kinetic energy of the system equals minus the asymptotic time average of its potential energy. However, while the kinetic energy is always nonnegative, the potential energy for repulsive coulomb forces is positive, contradiction.


