Is it possible to push together two charged particles of the same charge hard enough so it becomes a black hole? My chain of thought is the following:
To push together two charges of the same sign you need to do work. The energy spent will be turned into electrostatic potential energy.
Can we pump so much energy into the two particle system this way so it becomes a black hole?
 A: In principle yes, though in practice it's far beyond our current capabilities.
The problem is that a charged black hole has to have a Schwarzschild radius greater than $2r_Q$, where $r_Q$ is given by:
$$ r_Q = \sqrt{\frac{Q^2 G}{4\pi\varepsilon_0 c^4}} $$
If this isn't true the black hole will be superextremal and this (probably) means it's unstable. For two electrons colliding the charge will be $2e$, so the condition becomes:
$$ r_s = \frac{2GM}{c^2} \ge 2\sqrt{\frac{4e^2 G}{4\pi\varepsilon_0 c^4}} $$
or:
$$ M \ge \sqrt{\frac{4e^2}{4\pi\varepsilon_0 G}} $$
and the lowest vaue of $M$ works out to be about $3.7 \times 10^{-9}$ kg or about $2 \times 10^{27}$ eV. So you'd need to collide your two electrons with a centre of mass energy of at least 2000 yottavolts to get a stable black hole. It'll be a while before we get round to building an accelerator that powerful.
Even if we did have such a powerful accelerator there's another restriction to worry about. Unless we can line up the two electrons to within a small fraction of the Schwarzschild radius the resulting black hole would have a non-zero angular momentum, which would make it a Kerr-Newman black hole rather than Reissner-Nordstrom. We'd then have to worry about getting the angular momentum small enough for the KN black hole not to be extremal. I'll leave it as an exercise for the reader to work out how precise the alignment has to be.
