Aside from everything that was said by the others I would like to lay down the theoretical framework for a generalized solution (any speed, any mass, any charge, any distance, as long as the "balls" don't fall into a singularity).
There are 2 ways of looking at this problem.
The easiest one is to choose special relativity if the masses of the charges are relatively small, in which case we can neglect the gravitational effects. In such a situation we can use
$$ \frac{d(m_{0}U^\mu)}{d\tau} = - e (F'')^\mu_\nu U^\nu $$
with $$ (F'')^\mu_\nu = \Lambda^\mu_\alpha(-v_{rcv}) \Lambda_\nu^\beta(-v_{rcv}) (F')^\alpha_\beta $$ and $$ (F')^\mu_\nu = \Lambda^\mu_\alpha(v_{src}) \Lambda_\nu^\beta(v_{src}) F^\alpha_\beta $$ (for the rest of equations, what is which, how to combine them, see here)
We use this eq for each of foam-balls, and then we solve (using retarded positions) "orbits" at speeds close to the speed of light.
We vary $q_1$, $q_2$,draw graphs, deduce what happens.
Of course we need to define some border (surface) conditions, that are very important because they define what happens when the 2 balls collide.
Will they scatter ? Will they combine ? Will charge density combine to form peculiar new (kind of) matter ? Will they annihilate to create lots of EM waves or other kind of radiation ?
That is why it is very important to have the proper definition of what is under the exterior surface of the ball.
Simply assuming that is just a singularity under the exterior surface may disagree when it will came to compare with real life experiments.
The other way is to use general relativity.
There are two paths we could take here.
The simpler one is to assume that one of the ball has the charge and mass far far smaller than the other one : $m_1 >> m_2$ and $q_1 >> q_2$.
For such a case @Void provided here an answer in the framework of Reissner-Nordström metric, but I will try to answer from a bit different perspective the one of the theory of the bridges.
<PERSON> derived the metric in case of spherical symmetry for combined electrity and gravity a little bit different; he choose the sign of the energy tensor in such a way that by solving the field equations we obtain the metric $g_{\mu\nu}$ :
$$ ds^2 = (1 - \frac{2m}{r} - \frac{q^2}{2r^2}) dt^2 - \frac{1}{1 - \frac{2m}{r} - \frac{q^2}{2r^2}}dr^2 - r^2(d\theta^2 + \sin^2{\theta} d\phi^2) $$
So for such a metric the event horizon will be defined at $$\left(1 - \frac{2m}{r} - \frac{q^2}{2r^2} \right) = 0$$ This means that even without the help of mass we can get an event horizon.
Since we used the square of charge implies that it does not matter which sign has the charge.
For this using traditional black hole analysis we will came to the conclusion that anything that passes the event horizon will have no way to get out, no matter how close to the speed of light we go.
On the other hand <PERSON> suggested a change of variable that would help us get rid of the singularity of the event horizon. 298