# Can you electrically charge a black hole so that it would repel electrons?

Assume we somehow drop LOTS of electrons into a black hole, without the black hole accreting any other matter or energy at the same time.

I suppose the charge of the black hole would build up much faster than its mass. Would such a black start repelling electrons at some point, because the electric repulsion would become stronger than gravity?

If yes, would't it mean that the event horizon got smaller, because an electron reaching the original event horizon (before we started dropping lots of electrons into it) would now be repelled electrostatically instead of falling in?

• Related: physics.stackexchange.com/q/450380, but not a duplicate, because that question asks about black holes, and this new question doesn't require the infalling objects to be black holes – Chiral Anomaly Feb 8 at 19:00
• Also related: physics.stackexchange.com/q/12899/206691 – Chiral Anomaly Feb 8 at 19:07
• ...but also not quite a duplicate, because this new question asks about the event horizon getting smaller, which the older question (12899) does not address. – Chiral Anomaly Feb 8 at 21:13

This answer uses classical general relativity and ignores quantum effects, so the electron is treated as a classical pointlike particle. Otherwise, we'd need to use a quantum theory of gravity, which is currently too difficult.

I suppose the charge of the black hole would build up much faster than its mass. Would such a black start repelling electrons at some point, because the electric repulsion would become stronger than gravity?

Yes. For convenience, we can use units of mass $$M$$ and charge $$Q$$ such that two objects with $$M=Q$$ neither attract nor repel each other overall. The charge-to-mass ratio of an electron is $$q/m\sim 10^{21}\gg 1$$. If we start with an uncharged black hole with mass $$M_0$$ and then somehow manage to add $$N$$ electrons' worth of mass and charge, the resulting mass and charge would be $$M=M_0+Nm \hskip2cm Q=Nq, \tag{1}$$ where $$m$$ and $$q$$ are the mass and charge of an electron. Now consider the interaction between this black hole and a single distant electron. Using Newtonian gravity, the ratio of electrostatic respulsion to gravitational attraction would be $$\frac{Q}{M}\times\frac{q}{m} =\frac{Nq}{M_0+Nm}\times\frac{q}{m} =\frac{x}{1+x}(q/m)^2 \tag{2}$$ with $$x\equiv Nm/M_0$$. In this approximation, if $$x\gg (m/q)^2\sim 10^{-42}$$, then the electron will be strongly repelled because the ratio (2) is much greater than 1, even if we still have $$x\ll 1$$. This is true even when the electron is far enough away for the Newtonian-gravity approximation to be an excellent approximation, so this approximation is justified in hindsight. This approximation isn't good enough to tell us what the threshold value of $$N$$ would be, but it is good enough to confirm that electrons will be repelled if $$N$$ is large enough.

wouldn't it mean that the event horizon got smaller...?

I don't have a definitive answer. Here's what I do have. According to [1], the area of the event horizon of a charged (Reissner-Nordström) black hole with mass $$M$$ and charge $$Q$$ is $$A=4\pi R^2 \hskip2cm R \equiv M+\sqrt{M^2-Q^2}. \tag{3}$$ If $$Q>M$$, then the event horizon is absent, exposing a naked singularity. According to equations (1) and (3), after adding $$N$$ electrons to an initially-uncharged black hole of mass $$M_0$$, the area of the resulting event horizon would be $$4\pi R^2$$ with $$R=M_0(1+x)\left( 1+\sqrt{1-\frac{x^2}{(1+x)^2}(q/m)^2} \right) \hskip2cm x\equiv\frac{Nm}{M_0}. \tag{4}$$ This function $$R(x)$$ is a product of two factors, the factor $$1+x$$ being an increasing function of $$N$$ and the other factor being a decreasing function of $$N$$. The area initially increases with $$N$$, but beyond some threshold value of $$N$$, it begins to decrease. The question is whether or not this threshold is reached before further electrons start to be repelled. The analysis is not easy, for two reasons:

• Newtonian gravity is not a good enough approximation in this case.

• We can't use Hawking's area theorem, because that theorem assumes that naked singularities are absent [2], and the electron (treated as a classical pointlike particle) is a naked singularity because $$q/m\gg 1$$.

The closest analysis I found in the literature is the paper [3]. That paper considers a black hole that is already almost maximally charged ($$Q\lesssim M$$) and shows that we cannot make the horizon disappear ($$Q>M$$) by adding charged matter with $$q/m>1$$. Intuititively, this is because if $$q/m$$ is large enough to make the horizon disappear, then the matter will be repelled rather than attracted; and if we give it enough of a push (enough kinetic energy) to overcome this repulsion and force it into the black hole, then we have increased its energy enough so that $$q/E$$ is no longer large enough to make the horizon disappear. This doesn't quite answer the question, because the question is whether or not the horizon can be made smaller, and the paper [3] only explicitly asks whether or not we can make it disappear (and the answer is no). But the analysis in [3] is relatively detailed, and they also cite a couple of similar analyses, so with some effort, maybe those analyses can be adapted to answer the OP's question. The details are left as an exercise for the reader.

References:

[1] Section 8.6.5 in Straumann (2013), General Relativity (Second Edition)

[2] Page 6 in Düztas, "Cosmic censorship and the third law of black hole dynamics," https://arxiv.org/abs/1706.03927

[3] Sorce and Wald (2017), "Gedanken Experiments to Destroy a Black Hole II: Kerr-Newman Black Holes Cannot be Over-Charged or Over-Spun," https://arxiv.org/abs/1707.05862