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What prevents a static black hole from accumulating more charge than its maximum? Is it just simple Coulomb repulsion?
Is the answer the same for rotating black holes?
What I understand from the answers given so far, is that maximum charge is a moving target. You can add charge to a black hole but Coulomb repulsion guarantees that you will do so in a manner than will increase "maximum charge value". Is this correct?
Coulomb repulsion it is. Specifically, if a black hole has a lot of charge, then particles with a high charge-to-mass ratio will be repelled. Anything that falls in will contribute "more mass than charge," heuristically, keeping the charge-to-mass ratio of the black hole from getting too big.
There is a way of seeing this more explicitly with the Reissner-Nordstrom (RN) metric $$ ds^2~=~-F(r)dt^2~+~F(r)^{-1}dr^1~+~r^2d\Omega^2 $$ where the $F(r)~=~1~-~r_0/r~+~(Q/r)^2$, $r_0~=~2GM$ and $Q$ the charge in length units. The metric has two critical points $$ r_\pm~=~\frac{r_0}{2}~\pm~\frac{r_0}{2}\sqrt{\frac{4Q^2}{r_0^2}} $$ These are the outer and inner horizons for $r_+$ and $r_-$ respectively. The region between them is a spacelike trapping region, similar to the interior of a Schwarzschild solution. The extremal condition on the black hole is where $r_+~=~r_-$ which is where the spacelike region between the outer and inner horizons has been “removed,” or in a more subtle way mapped into the spacetime $AdS_2\times S^2$.
From the metric components we then compute the Christoffel symbols in the usual straight forwards, though tedious, manner. The most salient of the connection terms is $$ {\Gamma^r}_{tt}~=~F(r)\frac{r_0r~-~2Q^2}{2r^3} $$ which gives the geodesic equation $$ \frac{d^2r}{ds^2}~+~{\Gamma^r}_{tt}U^tU^t~=~0. $$ Far from the black hole We have that $U^t~\simeq~1$ and so $ds~\simeq~dt$ and this is a Newton second law type of equation $$ \frac{d^2r}{dt^2}~+~F(r)\frac{r_0r~-~2Q^2}{2r^3}~=~0, $$ where for $Q~=~0$ recovers Newton's second law for gravitation.
Now consider the extremal case. The connection term is then $$ {\Gamma^r}_{tt}~=~\frac{1}{2}\Big(1~-~\frac{r_0}{r}~+~\frac{r_0^2}{4r^2}\Big)\Big(\frac{r_0}{r^2}~-~\frac{r_0}{2r^3}\Big) $$ which tells us that a neutral particle is still attracted into the black hole. Then we consider a charged particle
The field strength 2-form and tensor components is $$ F~=~\frac{Q}{r^2}dt\wedge dr $$ The geodesic equation is no longer zero, but there is a driving force $F~=~F(r)r_0/2r^2$. With this Newtonian approximation the total force on the particle can be seen to be zero near the horizon. So for the extremal black hole a charge near the horizon will experience no net force.
Other connection terms are also nonzero. An important one is ${\Gamma^\theta}_{r\theta}~=~-1/r$ . For the extremal case the radial acceleration of a charge near the horizon approaches zero, but the angular component remains. Hence if there is a small $U^\theta$ this will move the charged particle off the radial path and ultimately away from the black hole. This in effect prevents the overcharging of a black hole.
the energy required to bind together N electrons in the sphere, it is actually
$e/R + 2e/R + ... Ne/R = N(N-1)e/2R$
which is quadratic.
For each additional electron shoot into the black hole we will have to add energy to it that is quadratic in the current charge (proportional to N, the existing charge), while the charge will just increase linearly (by one)
so the (additional) mass-energy of the black-hole will grow quadratically in the charge
since the maximum charge of a black hole is some function of the mass of the black hole, this implies that this function grow no quicker than a square root of the mass
The coulomb force cannot be responsible, because given enough energy I can add more charge to the black hole... There was a related question recently: Paradoxical interaction between a massive charged sphere and a point charge To summarized my asnwer there: It isn't actually the Coulomb force that prevents the addition of charge to the black hole beyond its maximum, but by adding more charge more electrostatic binding energy is added. A lot! (since the black hole must have an incredibly chargedensity.) Therefore also its mass increases, which in turn leads to an increased maximum charge.....
There is actually a nice and simple calculation of gravitational collapse of charged spherical shells, where you can show that Coulomb repulsion is stronger than gravitational attraction if you exceed the critical bound |Q|>M (in convenient units). You find this simple calculation in the lecture notes by Paul Townsend on black holes [see chapter 3, in particular eqs. (3.10)-(3.13)].
A slightly different answer to this is "neutralisation". That is the free positive ions around (in gas clouds nearby maybe) will neutralise the charge. This is generally assumed to keep the charge of a Black Hole near to zero in astrophysical contexts.
