Physical interpretation of the Reissner-Nordstrom metric for the special case In the Reissner-Nordstrom metric
 $$ \mathrm ds^2 = \left(1 - \frac{2GM}{r} + \frac{GQ^2}{4\pi\epsilon_0 r^2}\right)~\mathrm dt^2 - \left(1 - \frac{2GM}{r} + \frac{GQ^2}{4\pi\epsilon_0 r^2}\right)^{-1} ~\mathrm dr^2 - r^2~\mathrm d\Omega^2$$
 when we consider the case with
$$ \frac{2GM}{r} =  \frac{GQ^2}{4\pi\epsilon_0 r^2} \;,$$
what is the meaning of this case?
Does Electromagnetic force counterbalance the gravitational force?
 A: With a Reissner-Nordström black hole there are two contributions to the stress-energy energy tensor, the mass and the electrostatic field.
The mass is all at the singularity at $r=0$ so if we consider a sphere of any radius $r$ all the mass is still inside the sphere no matter how small we make $r$. So all the mass is in effect still pulling you inwards.
However the electrostatic field fills all of space because the field strength vaies as $1/r^2$. This means that for a sphere with some radius $r$ some of the field is inside the sphere and some is outside the sphere. In the RN metric it's the energy contained in the field that curves spacetime, which is why it affects neutral particles. So in effect some of the electrostatic field is inside the sphere and pulling you inwards while the rest of it is outside the sphere and pulling you outwards.
So at the radius $r$ where:
$$ -\frac{2GM}{r} + \frac{GQ^2}{4\pi\epsilon_0 r^2} = 0 $$
the mass plus the energy of the electrostatic field inside $r$ is exactly balanced by the energy of the electrostatic field outside $r$.
Note that it isn't the case that the gravity and the electrostatic force balance because the electrostatic force is the charge times the field strength. So if you have charged particles with the same mass but different charges the forces would balance at different points. The radius $r$ you've described is where the energies of the mass and the field inside and outside $r$ balance.
