TITLE: Movement of repelled particles in a ball
QUESTION [6 upvotes]: 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 $t>0$?

REPLY [7 votes]: If all the particles remained in a bounded domain, the 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.