1977ub said:
Does The Presence of Charge Add To Relativistic Mass-Energy, or does it take two particles attracting or repelling one another to generate charge-related mass-energy?
While PeterDonis has answered the question, you can see this at work in the
Reissner–Nordström metric which is the vacuum solution for a charged black hole. Gravity for a charged black hole is-
[tex]a_g=\frac{M}{r^2\sqrt{1-\frac{2M}{r}-\frac{Q^2}{r^2}}}[/tex]
where [itex]M=Gm/c^2[/itex] (mass in geometric units) and [itex]Q=C\sqrt(Gk_e)/c^2[/itex] (charge in geometric units) where [itex]G[/itex] is the gravitational constant, [itex]c[/itex] is the speed of light, [itex]C[/itex] is the charge in Coulombs and [itex]k_e[/itex] is Coulomb's constant (as a rule, [itex]M\geq Q[/itex]). Multiply the answer by [itex]c^2[/itex] for S.I. units of gravity (m/s
2).
If charge is removed and the black hole is reduced to it's irreducible mass (the same black hole becoming a Schwarzschild black hole) where-
[tex]M_{\text{ir}}=\frac{r_+}{2}\ \ \rightarrow\ \ M=\frac{Q^2}{4M_{\text{ir}}}+M_{\text{ir}}[/tex]
where [itex]M_{\text{ir}}[/itex] is the irreducible mass and [itex]r_+=M+\sqrt{M^2-Q^2}[/itex] is the outer event horizon, the equation for gravity becomes-
[tex]a_g=\frac{M{\text{ir}}}{r^2\sqrt{1-\frac{2M{\text{ir}}}{r}}}[/tex]
you'll notice a distinct drop in the gravity field at a specific r some distance from the black hole. It might also be worth looking at Kerr-Newman metric which also includes for spin. Up to 50% of RN black hole's 'mass' can be attributed to charge though charged black holes are not considered realistic.
More info-
https://www.physicsforums.com/threads/what-is-irreducible-mass.762993/