reissner nordstrom metric - Can a nearly-extremal black hole be stable against Schwinger vacuum breakdown?

I was doing some basic algebra to estimate the range of possible masses $M$ and electric charge $Q$ for a nearly extremal Reissner-Noström black hole. I want to see if the logic is correct

the electric field near the event horizon of such a black hole would be

$$ E= \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{q c^4}{G^2 M^2} = 10^{66} \frac{\text{Volt Kg}^2}{\text{m Coul}} \frac{q}{M^2}$$

since Schwinger-pair vacuum breakdown occurs at $10^{18} \frac{\text{Volt}}{\text{m}}$, it implies that

$$ \frac{q}{M^2} \lt 10^{-48} \frac{\text{Coul}}{\text{Kg}^{2}}$$

Would be a bound that guarantees electromagnetic vacuum stability

Question: can a charged black hole that is stable against this breakdown still be nearly extremal?

well, it seems so, because the extremality condition:

$$ r_Q = \frac{r_s}{2}$$

Implies that

$$ Q \sqrt{ \frac{G}{4 \pi \epsilon_0 c^4}} = \frac{GM}{c^2}$$

$$ \frac{Q}{M} \approx 10^{-16} \frac{Coul}{Kg}$$

So the above two relations would imply that a black hole with a mass

$$ M \gt 10^{32} Kg \approx 100 \text{ Solar masses}$$

Can be both stable (against the specific Schwinger-pair decay) and extremal.

In particular, if the mass is $10^{32}$ kilograms, a charge around $10^{16}$ Coulomb would make the black hole arbitrarily close to extremal as one might want, and is not at all clear that it would decay from the extremality state

But probably i'm overlooking other ways than the extremality of the black hole can decay (in particular, i'm ignorant about how Hawking radiation will contribute in here, specially if the temperature is below 511 KeV). Any insights into this would be greatly appreciated