A DC voltage source supplies electrons that move in a single direction, producing a current $I$ distributed through the conductor's cross-section.
As per Ohm's law, a change in voltage changes the force on the electrons, causing an increase in the speed (and hence the current) of the electrons.
Electrons under the influence of a DC voltage don't move in a straight path, as they are attracted by the positive charge of atomic nuclei in their path.
When we apply a sinusoidal AC voltage to a conductor, electrons alternately accelerate, slow, stop, and reverse direction.
When we apply high frequencies, like $300$ MHz, I assume that the alternation of voltage direction is so fast that electrons don't move, they only hesitate in one place. But if this is true, that electrons are standing still, then how do antennas oscillate current to create $EM$ waves?
If we look at the plot (below), with two different frequency waves, it appears that the electrons get to the same speed (or amplitude) with different frequencies, but they just reach it faster at higher frequencies.
What is the reason for the energy difference between the higher and lower frequency waves?
- Possibly the difference is due to the energy of a quantum of each $EM$ wave since its energy depends on its frequency $E=h\nu$?
- Possibly the difference is due to Special Relativity effects. Maybe the $E$ field of the oscillating electron increases and compresses with the increase in velocity of the electron?
- In my example (below), the frequency doesn't change the amplitude of the wave. Is it the change in electron speed that increases the EM wave energy? Or, is it the peak voltage upon which the EM wave energy depends?
Example - illustrating my question:
If I generate a $300$ MHz $EM$ wave through an antenna with a $1$ V voltage source, will that EM wave have higher energy than a much lower frequency wave produced by the same antenna but with a $200$ V amplitude? If so, why?