"Electromagnetic" Black Hole Suppose you take a large arbitrary positive electric charge $Q$.
Then, given a distance $r$, we can calculate the escape velocity $v$ necessary for an electron to escape the positive charge (just like a rocket trying to escape Earth's gravity).


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*Could this escape velocity be equal or greater to $c$?

*Would this create a space from which no negative charge can escape? A sort of "black hole" for negatively charged particles?

*Does this mean no positively charged particle could enter this space? Is this a sort of "white hole" for positively charged particles?

*Suppose that you "free" one of the positive charges inside. The electric force would obviously repel it incredibly strongly. I seem to remember something about that the speed of particles falling into a black hole approaches the speed of light. Would a positive particle escaping from our "white hole" appear to be moving at the speed of light, at the event horizon?


I tried to make some back-of-the-envelope calculations, but without taking relativity into consideration I don't really trust the results.
 A: There are charged Black Holes, with stationary solutions known for both spherical symmetry and axial symmetry with rotation. Those may be the only stationary Black Holes (BH's) possible, by the No Hair theorems (they are for spherical symmetry if you assume a positive energy condition, and possibly other constraints, and I can't remember if also unique for rotating holes, with similar constraints)
Regardless of the uniqueness, which may be so except for scalar or maybe other quantum fields, the important thing about charged BH's is that there's a limit to how much charge they can have. If they have too much, compared to the BHBH mass, the gravitational attraction won't be enough to hold the mutually repelling charges in. It is not that a BH will break part, that can't happen, it's that they will never form with too much charge. It's been simulated and theoretically studied, and charges just don't go inside the horizon when there is too much already there. 
The same is true for angular momentum. There's a limit, and if you try to drop a high angular momentum particle on a BH already at the limit, it'll just have enough angular momentum to fly off and not make it in. 
See http://www.physics.umd.edu/grt/taj/776b/chappell.pdf. The bottom line is that a BH with $m^2$ less than ($Q^2 + a^2$), in natural units and with Q the total charge and 'a' a measure of angular momentum is not possible. Some simulations verify that, but it is not fully proven. It is known that in those cases the horizons don't form, and if there is any singularity then it'd be naked singularities. The belief is that it is not possible. 
