I still don't think you're "getting" a critical thing.
The voltage on the gate is actually irrelevant.
It's the voltage on the gate *WITH REFERENCE TO THE VOLTAGE AT THE SOURCE* that will turn the FET on.
So, if you have 50 volts at the rectifier output (point (C) on your diagram), the drain will be at 50V.
The source will be at (lets say) 12V, when the FET is off.
So your charge pump makes 24V. OK. Fine, you hit the gate with that 24V, and at that instant Vs=12, Vg=24, Vgs=12 and the fet starts to turn on.
AS it turns on, the 50V on the drain starts appearing on the source. So it is now ramping from 12V up "towards" 50V. In super-slow-motion... source starts getting power through, it's now at 13V. Still 24V on the gate, Vgs now 11V. FET still switching on, source now 14V, gate still 24V, Vgs now 10V, FET still switching on, source now 15V, gate still 24V, Vgs now 9V, FET may still switching on but not so hard... source now to 16V, Vgs only 8V.... at 20V on the source, Vgs is now only 4V - you see what's happening??
This is ALL because you're driving the gate with a ground-referenced signal, not a source-referenced signal.
The design is fundamentally flawed, and NO amount of fluffing about with it will make it work "properly"
Similarly, turning it "off", as has already been stated - a 1 megohm resistor will bleed charge off quite slowly. It certainly won't be DRIVING the FET off. There's a reason this type of circuit isn't used in production. It just doesn't work.