How can two charged black holes merge despite electrostatic repulsion?

Question

I have read this question:

Collision of charged black holes

And it made me curious.

I understand that the charged black holes do have negative EM charge, and they repel.

This EM interaction and repulsion between electron fields around nuclei, cause matter to have spatial extent. This EM repulsion is why atoms can't get closer to each other then a certain distance. And why atoms in molecules can't get closer to each other then a certain distance.

This is the reason why matter is 99% space.

Of course, the Heisenberg Uncertainty principle has an effect on this too.

Question:

1. How can two charged (negative EM charge) black holes merge? How can gravity overcome the EM repulsion and the Heisenberg uncertainty principle?

2. Do these electron fields pass through each other (merge too) when the holes merge? Does the Heisenberg uncertainty principle and the Pauli exclusion principle for fermions not apply anymore?

Answer 1

Two charged black holes are very much like two charged droplets: they can merge, as long as the charge is not strong enough to repel them. The result is a bigger hole with the sum of the charges. It is not like the EM repulsion goes to infinity as they approach each other: they have a spatial extent, and the surface electric fields will be roughly following Coulomb's law (with some corrections due to curved spacetime).

Approximately, two black holes of mass $M$ and charge $Q$ each will repel each other if $GM^2/r^2 < kQ^2/r^2$, or $Q/M > \sqrt{G/k}\approx 10^{-10}$ C/kg. For a solar mass black hole that is about $10^{20}$ C. This produces a field at the horizon that is way past the Schwinger limit where the electromagnetic field becomes nonlinear, so it is likely that long before this the whole system destabilizes in a shower of electrons and positrons.

The Heisenberg uncertainty for macroscopic black holes is so small that it is negligible. Whether black holes can be treated as fermions looks uncertain.