I opened this thread thinking that The Tropical Isle had just invented another mixed drink.
A punk bossa nova band.
Hmm. I was hoping the high ratio between the electrostatic force and the gravitational force would be more than enough here, but the black hole is so small that the self-energy of the charge is probably not negligible. On the other hand, the answer also depends on how much you can charge up the base plate. That will have its own limit, and you would have to stop before that becomes a black hole as well.
If my math is right, my 19 gigatonne black hole could simply be a ball of 2.97 coulombs of charge packed into a ball with radius 2.8e-17 m. That’s a surprisingly ordinary amount of charge (considering that most of the numbers here have exponents of >10 or <-10).
And that is a bit of a problem. The black hole weighs 1.85e11 N. To levitate it 1 m requires a base with charge ~20.5 C. And that is quite a lot of charge indeed. It doesn’t need to be packed into quite the same volume, but even a moderately small volume will require tremendous energy. Charged particles will really want to shoot off in all directions. And that’s only levitating one meter; hardly enough space to protect against the black hole that’s radiating a terawatt of power. You’d really want kilometers of space here.