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If there is a strongly charged black hole, and a particle with a strong charge of the same sign goes inside of the black hole's event horizon while moving quickly, is there a reason why it would not be able to come out the other side of the black hole's event horizon?

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No.

The reason: Below the event horizon, there is no outgoing timelike direction. It means, it doesn't matter how do you accelerate a particle, it will move inwardly:

Causal cones in the BH

The cones are the light cones of the object, what means if it would send out a radio signal in every direction, it would go in these cones. The best reachable orbit (-> the most delayed crash into the singularity) is also on the mantle of these cones.

It is stronger thing as you would imagine it in a Newtonian viewpoint. It is not that the "BH pulls the particle too strongly", what you could maybe equalize with a more strong push. It is like the BH curves the space(time) in such a way, that literally there is no way out.

Maybe it would be a more interesting question, if we could leave the BH with another, smalling BH falling into it. It would be a deviation from the Kerr (Schw, Nordström, etc) metric, in which maybe a timelike outgoing orbit would be possible.

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    – peterh
    Commented Jul 25, 2016 at 1:13
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Once in it won't come out, it just becomes part of the BH, adding mass and charge (plus or minus), and probably angularity momentum, to the BH. When something gets inside the horizon, it can never escape, out of any side of the BH.

There is another process where one can use charged particles to extract energy from the electric energy of the BH. This is similar to extracting energy from the angular momentum, except in this case it's the charge. It can be a combination. The paper below derived the charge case. The idea is the gravitational field augments the charge of the incoming particle, as a result extracting energy and charge (and probably angular momentum) from the BH. Originally the idea came from Penrose for extracting the energy from the angular momentum of the BH, called the Penrose process.

http://journals.aps.org/prd/abstract/10.1103/PhysRevD.7.949

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