The fact that you can extract only half the power from a wave with a
perfectly efficient antenna is already built into Ae. That is, Ae = P/S,
where P is the power delivered to a conjugate load and S is the power
density of the wave.

I suspect that in your reference, Ea is the *peak*, not RMS, field
strength, and that's why the factor of two.

Incidentally, while you can extract only half the power from a wave with
a perfectly efficient antenna, it is possible to extract *all* the power
from a wave. An anechoic chamber extracts essentially all the power from
waves generated within it.

Roy Lewallen, W7EL

Ron wrote:
What is the available power, Pa in picowatts, that can be extracted from
a free space field of Ea microvolts per meter?

A reference I have says it is:

Pa = Ea squared multiplied by Ae (the antenna effective area) all
divided by 2Zo, where Zo = 377 ohms, the impedance of free space.

My question is why 2Zo rather than just Zo. Is it because only half of
the power in the field can be extracted and delivered to a load?

Ron, W4TQT