In a general case P = V^2 / R

That said, at the manufacturing scales of these parts leakage in the transistor can be as big a power consumer as the switching.

Leakage current increasing as voltage increases at a rate that is of the order ~e^Vth

I wouldn't really say its strictly either exponential or quadratic, you'd have to know a lot more about the process and the implementation to know how significant leakage current is in the design.

Yes you are correct, the power used to switch a transistor is P = a * V2 * C * f.

The static power dissipation from leakage currents though is Ileak = Isat * (e(qV/kT)-1).

http://www.siliconintelligence.com/people/binu/perception/no...

Dynamic power is quadratic with voltage and is work-load dependent, i.e P = c * AC capacitance * Freq * V^2. Static (i.e. leakage) power has an exponential component (P = AVe^(k*V)), although k is typically pretty small. I'm not sure about the static vs. dynamic power breakdown of Apple's designs.