Why can the energy of local computing be written as \$f^2\$,not \$v^2\$
We can divide the losses in FET logic (all processors are made in FET logic) in to categories:


static losses, i.e. leakage currents,
switching losses.


The reason is simple: Because they use FETs, the transistors don't need any current to flow through their gate to control the output. Therefore, the transistors use no current at all – aside from leakage.

When switching, however, the charge in the gate capacitor of a FET has to be changed – which means a current needs to flow. Since resistances are non-zero, with P=V·I and Ohm's law, it follows that P = I²·R.

To switch faster, i.e. to have a higher clock frequency, you need to have a higher current flowing in (simple: a gate capacitor exposed to a higher voltage charges faster, like every other capacitor; current is amount of charge per time) or out the gate capacitors every clock cycle. Therefore, I is (at least!) proportional to f, I = µ·f.

Therefore, P = I²·R = µ²·f²·R.

µ and R are material/structural constants of your semiconductor technology (this is a bit simplifying, but it doesn't really matter whether losses are purely ohmic or also have higher potences of the voltages involved here). 

Therefore, P is proportional to the square of frequency, at the very least.

That's why the Pentium IV generation of processors, designed to clock incredibly fast, was extremely power-hungry.