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 $t>0$?