 Solution

For $M>0$ and $Q^2>2M^2$, the physical exterior starts at $r=b>2M$. Throughout it, $f>0$, so the putative zero of $f$ is not in the exterior and no regular horizon shields the curvature singularity. The normal to $r=b$ has positive limiting squared norm $g^{rr}=f(b)>0$, identifying a timelike singular boundary. Outgoing radial null geodesics obey $dr/dt=f(r)>0$ and escape from arbitrarily close to that boundary. HenceAt $Q^2=2M^2$, the singular boundary coincides with $f=0$ and is null. There is no regular shielding horizon of positive area; this marginal singular extremal case must be separated from both the regular black hole and the supercritical timelike naked singularity. If “naked” is used simply to mean unshielded by a regular horizon, equality belongs on that side of the distinction, but its causal boundary is different.

The usual positive-mass restriction matters. For $M<0$, $f(r)>0$ at every positive radius and there is no positive-radius horizon; the central zero-area boundary at $r=0$ is singular and naked. For $Q=0$ its angular sectional curvature is $2M/r^3$, which diverges there. For $Q\ne 0$, $C=r(r-b)$ with $b<0$, and the angular sectional curvature $[1-f(C^{\prime }{)}^2/(4C)]/C$ also diverges as $r\downarrow 0$. The displayed family is undefined at $M=0$ with nonzero $Q$; its uncharged zero-mass limit is flat.