Reg`s question was:
"What is the voltage between the bottom end and ground of a 1 metre high
vertical antenna, above perfect ground, when the vertically-polarized
field strength is 1 volt per metre, and the antenna height is shorter
than 1/4-wavelength?"

A very clear and succinct question, I thought.

Roy Lewallen wrote:
"Relative to what?"

Reg left little to assume. I inferred that Reg had meant a
ground-mounted 1-meter whip on a small base insulator. That would have
left a short distance between the points of voltage determinarion.

The 1-meter whip directly over flat perfect earth must have a conjugate
match to its receiver to extract all available power and get maximum
voltage at the receiver input. This requires a low-loss coil to tune out
the high capacitive reactannce of a too-short whip. It was specified as
being less than 1/4-wavelength.

Standing wave antennas must be resonant to allow maximum current flow.

An unbalanced whip antenna must consist of two electrical parts just as
a balanced antenna does. The ground or ground plane used with a whip
substitutes for the missing half of a dipole. It provides a virtual
image of the whip above it, to complete a dipole-like antenna. In this
instance, the ground surface intervenes splitting the antenna into real
and virtual parts. Radiation comes from the real part above ground. This
part has only half the impedance of a totally real dipole.

We don`t need to have problems in determining the base voltage of a whip
antenna. We can measure the antenna`s base impedance with a bridge, and
its current with a thermoammeter with good accuracy, then we can
calculate the voltage at the base of the whip.

The answer to Roy`s question is: The r-f voltage at the base of the
antenna is determined with respect to ground at the base of the antenna,

Best regards, Richard Harrison, KB5WZI