IMGOOD, I see what you're saying.
However, I think everything isn't always as simple as Ohm's law, or we are viewing the problem the wrong way.
Ultimately, the static potential is just that - a potential, with little to no capability to "pump" a lot of current.
However, a 10kV generator, spun by a large mechanical prime mover, can pump out lots and lots of amps (and thus V*I=power).
I don't know if these are exactly realistic numbers (atleast for the path resistance), but let's assume the following:
Let's assume you are holding your hands across the armature windings of a giant, spinning,dc generator ( I wouldn't recommend this), and compare to a static discharge when moving from your desk:
Vstatic=10 KV
Istatic= 0.0001 A (negligble amount of amps the static discharge can source)
Vgenerator=10 KV
Igenerator(max)(due to winding resistance limitations, prime mover limitations etc etc)= 500 A.
Rgenerator(armature)=10 Ohm
Rpathofhuman= 990 Ohm, Rpath_effective=Rgenerator+Rpathhuman=1000 Ohm
So, the static electricity can only source 0.0001 A. So, the "power" dissipated by you when you move from the chair is:
P=V*I= 0.0001 A * 10 kV = 1 w (it doesn't matter that I=Vstatic/Rpath=10A, there's no way a static discharge is going to source 10A @ 10,000 kV , else everyone would be fried by standing up from their desk.. :) )
The generator, could source up to 500A. However, since Reffective=1000 Ohm, it will only source, depending on the speed (back emf) (assume low):
I=V/R= 10kV/1000=10A
So, P=V*I = 10,000 V * 10A = 100,000 W (i.e. you are toast)
I'm not sure if this is a good analogy, someone correct me if I'm way off base.
I'm not really willing to do a real experiment on it :)