Roy, W7EL wrote:
"If the wavelength is 1m, the voltage induced in the center of an
open-circuited 1m dipole by a 1V/m field is 0.5 volt, not 1 volt."

An open-circuited dipole (has a small gap in its middle) is mot
resonnant at a wavelength of 2 meters, but its individual pieces are
resonant at a wavelength of 1 meter. At 2 meters, not a wavelength
1m, each 0.5 m piece of the half-wavelength, 1 meter long dipole has a
high reactance because 0.5 m is too short at a wavelength of 2 meters
to be resonant. At longer wavelengths, the reactance rises.. High
reactance does not oppose non-existing current in an open-circuit.

I agree the voltage induced in 1/2-meter of wire properly placed within
a 1V/m uniform field is 0.5 volt, not 1 volt. The induced voltage in a
wire within a uniform field sweeping the wire rises uniformly along the
wire. It can be assumed to be the summation of tiny increments of
voltage all along the wire. The voltages of the too-short dipole halves
add just as two cells in some flashlights add. Their vectors are head to
tail. But, current will be limited by radiation and loss resistances of
the wires. It will also be limited by reactance in the wires.
Open-circuit, 0.5 V + 0.5 V = 1V.

Best regards, Richard Harrison, KB5WZI