Pumping charged particles (of same charge) into a blackhole Would would happen if you started pumping charged particles of same charge into a black hole?  Let's assume that you have an infinite number of those charged particles.  What will happen to the event horizon and the singularity?  Please give both perspectives - that of the charged particle's falling in and that of an external observer's.
 A: As stated, the charged black hole is a solution in general relativity known as a Reissner-Nordström black hole. Since you're looking for more of a conceptual overview, I'll skip the complicated maths and go straight to the end interpretations.
Let's give our black hole the electric charge, $Q$, the magnetic charge, $P$, and a mass, $M$. With this, there are three possibilities for what this black hole looks like:


*

*$Q^2+P^2>GM^2$ where $G$ is Newton's gravitational constant. In this case the charge is much greater than the mass and we find that there is no event horizon. The black hole becomes a naked singularity. However, this is unphysical, mostly because nothing would hold all the charges to the black hole but also for more mathematical reasons. So there's no point going further into detail about something that doesn't resemble physical nature.

*$Q^2+P^2<GM^2$. This solution is described as being "wormhole-esque". The event horizon of a regular black hole is such a problem because crossing it turns surfaces of constant radius into spacelike trajectories. That means you can only continue forward and to hold position above the singularity requires going faster than $c$. However, for this solution, we find that the black hole has 2 real event horizons. The outer horizon acts like a regular one. When you cross it, you are doomed to fall through the inner horizon. The inner horizon has a sort of cancelling effect. When you cross the inner horizon, surfaces of constant radius once again become timelike, which means you can choose to remain still, hit the singularity, or exit through the inner horizon again. If you go back through the inner horizon, things flip again and you become doomed to cross the outer horizon (you must keep moving forward). However, this does not return you to where you started from. The universe you return to is a copy of your original spacetime, but it is not the original.

*$Q^2+P^2=GM^2$. This is much the same as case (2) except that the outer and inner horizon have the same radius. It would act similar to a regular event horizon from the outside, except that there's no region where surfaces of constant radius are spacelike. This is as close to a naked singularity as one can get.
A: Well if you add charge gradually to a black hole, it will become more charged. If you started with a neutral Schwarzschild black hole, it will then become a Reissner Nordstrom one, and each time you throw in another charged particle the charge will increase. 
As the charge of a black hole is increased (keeping it's mass fixed), it becomes colder (if you are unfamiliar with the fact that black holes have a temperature, see this Wikipedia article: http://en.wikipedia.org/wiki/Black_hole_thermodynamics). So eventually the black hole will become very cold and then reach a zero-temperature extremal black hole.
Except that's not physically realizable, because all the charged particles we have at our disposal also have mass. So electrons are thrown in, the mass of the black hole is increasing and so is the charge. It turns out that this fact means that the black hole will never reach zero temperature, but will just get colder and colder with each new electron. From a more ordinary thermodynamics point of view, this is pleasing, since you should never be able to reach $T=0$ in a finite number of steps.
