It may be tempting to imagine that, as the spacetime descrbing the region between an observer suspended just above the event horizon and the event horizon itself may have negligible curvature, that such an observer will measure an infalling object to cross the event horizon in finite time.
However this is not so, the key thing perhaps to note is that whilst the small region of spacetime may be so flat that we could model it as Minkowski spacetime without fear, the worldine of the observer suspended just above the horizon will always be very curved.
If we take the observer suspended above the black hole to be a Rindler observer in flat spacetime, the even horizon now becomes the Rindler horizon.
If you look at the relationship between Minkowski and Rindler cooridinates then you will see as x->0 (x = 0 is the Rindler Horizon), then for the Minkowski cooridnates T and X, T/X ->1, therefore Rindler time t = arctanh(T/X) -> ∞. I.e. for our Rindler observer (for whom the Rindler time t is their proper time) events at the Rindler horizon always lie in the infinite future.
edited to add: I’m not sure though of your reasonign behind an observer supsended just otuside the hole not being able to observe the em field. In the usual course of events, they would be able to.
For a faraway observer dr/dt = -(1-2M/r)(2M/r)[sup]1/2[/sup] from which we find the velocity of the infalling charged sphere to go to zero at r = 2M the horizon.
For a suspended observer dr/dt(suspended) = -(2M/r)[sup]1/2[/sup] from which we find the velocity of the infalling sphere to go to 1 or c at r = 2M the horizon.
So for the suspended observer the falling object passes across the horizon with the speed of light.*
directly from John Wheeler in Exploring Black Holes pg.3-15
Integrating dt = dr/-(2M/r)[sup]1/2[/sup] from 3M to 2M appears to come out as a finite time.
However I’m certainly no expert of this stuff and I most certainly could be totally wrong.
Also, I feel I must say that I don’t see how my above post can be correct. The SR length contraction for the infalling sphere exactly offsets the space curvature so I can see how on the the sphere’s clock it will cross the horizon in a finite time.
But it doesn’t seem like that would be true for the suspended observer’s proper time.