With how much energy can a charged black hole launch a electron away with? Lets say with have a black hole with a strong negative charge, then a say a neutron falls towards it.
Just before the neutron reaches the black hole's event horizon it undergoes beta decay, this further pulls the proton towards it, while launching the electron away.
Exactly how energetic can this electron get?
 A: In general relativity, the solution of the Einstein-Maxwell equations for a non-rotating charged black hole is called the Reissner-Nordström solution (references 1,2). It has an electrostatic field in addition to a gravitational field. Far enough away from the event horizon, both effects are well-approximated by the familiar $1/r^2$ force laws. So, to keep the math easy, suppose that the beta-decay occurs far enough away from the black hole that the $1/r^2$ force laws are good enough approximations for both gravity and electrostatics, and suppose that the electron's initial velocity is zero. Then the electron's final kinetic energy will be the same as its initial potential energy, as defined using those approximations.
The magnitude of the gravitational potential energy in natural units is $mM/r$, where $m$ is the electron's mass, $M$ is the black hole mass, and $r$ is the radial coordinate of the electron, defined so that the area of a spherical surface centered on the black hole is $4\pi r^2$. (I defined it that way because the naive definition "distance from the center of the black hole" is meaningless.) The magnitude of the electrostatic potential energy in natural units is $qQ/r$ where $q$ is the electron's charge and $Q$ is the black hole's charge. Assuming that the net force is repulsive, as it will be if the charge is large enough, the magnitude of the electron's final kinetic energy is $|qQ-mM|/r$.
(Throughout this answer I'm assuming that the black hole is large enough that it doesn't quickly neutralize itself as a result of Schwinger pair-creation, a quantum effect that occurs in sufficiently strong electric fields. Another answer by A.V.S. gives more information about this caveat.)
Now consider an extremal black hole, which means that its charge $Q$ and mass $M$ are equal to each other (in natural units). As explained elsewhere (like the end of my answer to another question), this is apparently the maximum amount of charge that a black hole can have. For something like a solar-mass black hole, this is an immense amount of charge. The net force on the electron is overwhelmingly repulsive, and its final kinetic energy has magnitude $|q-m|M/r$. The electron's charge is much greater than its mass (in natural units), $q\gg m$, so this is essentially the same as $|q|M/r$.
To ensure that the approximations we're using are adequate, we should take $r\gg M$, say $r=100M$. From that distance, the black hole would "look" roughly $2$ degrees wide, which is roughly four times as big as the moon looks from the earth. Then the electron's final kinetic energy has magnitude $|q|/100$.
How much kinetic energy is that? To find out, let's convert it back to Standard International units. The relevant constants are the gravitational constant $G$, the vacuum permittivity $\epsilon_0$, and the speed of light $c$. Up to a factor of order $1$ that we could ignore for this estimate (but I'll try to get it right anyway), the unique combination with units of energy is
$$
\frac{|q|}{100}\times \frac{c^2}{\sqrt{4\pi \epsilon_0 G}}
\sim
1\text{ million Joules}.
$$
That's roughly the amount of kinetic energy that a thousand-kilogram boulder has after free-falling from a height of a hundred meters to the surface of the earth, and we've packed all that energy into the motion of just one electron.
And remember, that's an underestimate, because the electron started at rest at a fairly large distance from the black hole. We could do better by letting the electron start closer to the black hole, but that would require a more difficult calculation because the approximations used above would no longer be adequate.
By the way, as another illustration of just how immense the charge $Q=M$ is, consider the Majumdar-Papapetrou solutions in general relativity (references 3,4). These solutions describe two or more extremal black holes, and they show that the net force between extremal black holes is zero if their charges have the same sign. The electric field is intense enough to counteract the gravitational attraction between two black holes, and when you put an electron in an electric field that strong, it gets quite a push!


*

*Chapter 3 in Townsend (1997), Black Holes (https://arxiv.org/abs/gr-qc/9707012)


*Chapter 31 in Blau (2001), Lecture Notes on General Relativity (http://www.blau.itp.unibe.ch/newlecturesGR.pdf)


*Hartle and Hawking (1972), Solutions of the Einstein-Maxwell equations with many black holes (https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-26/issue-2/Solutions-of-the-Einstein-Maxwell-equations-with-many-black-holes/cmp/1103858037.full)


*Reference 2, equations (31.38)-(31.39) and the accompanying text
