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While scrolling through an online document, I came across the statement, "If charges carried by the black hole are large, the curvature at the horizon is small." How is this so?

I tried to think of an answer to this. My thinking goes this way: I know curvature is related to mass. If an object is more massive, it bends the spacetime fabric more, producing large curvature. So curvature small means mass small. So how does a large charge mean a small mass for a black hole? What is the relationship between the charge and mass of a black hole?

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  • $\begingroup$ You may be interested in the Wikipedia entry for charged, non-rotating black holes (for simplicity). It explains that "the electric field energy also contributes to the total mass." $\endgroup$
    – Jeanbaptiste Roux
    Commented May 25, 2023 at 7:05
  • $\begingroup$ Does this answer your question? Physical interpretation of the Reissner-Nordstrom metric $\endgroup$
    – John Rennie
    Commented May 25, 2023 at 7:16
  • $\begingroup$ Usually when you use the various definitions of mass for charged BH, you will find that even the electric charge of BH will contribute to its final mass $\endgroup$
    – KP99
    Commented May 25, 2023 at 11:34

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