The "impedance" of free space/ air is 377 ohms. Thus the ratio of RF field
voltage and current will be
fixed and related to the power density - unless the field is distorted.

As another crude approximation, assume that (unless you know a radiation
pattern) that the unwanted signal's power is equally distributed over a
sphere or maybe a hemisphere.

R. J. & Ed-

The calculations R. J. refers to apply to a field at a distance - called "far
field". Up close, radiation patterns are complex and not easily predicted.

For the sake of discussion, lets carry this free space impedance topic a bit
further. If you know transmitted power and the type of antenna, you can
approximate the effective radiated power. For example, a half wave dipole and
a quarter wave ground plane have about the same gain, 2.1 dB. If you are
concerned about worst-case, you should assume a handheld radio with a
rubber-duckie antenna, has the same gain as a full size ground plane. (You
could also add some gain due to reflecting surfaces, just to be safe.)

Suppose your handheld radio has five watts output. For a gain of 2.1 dB, the
effective isotropic radiated power is about 8 watts. This power illuminates
the inner surface area of a sphere equal to (4/3) Pi R-squared, where R is the
distance.

At some distance the field strength will equal an amount that will interfere
with the defibrillator. Suppose that critical field strength is one Volt per
Meter. One Volt squared, divided by 377 equals 2.7 milliWatts per square
Meter.

Working this backwards with the sphere, 4/3 Pi R-squared divided by 8 watts
equals the power (2.7 mW) per square meter at distance R. You can calculate
that the minimum distance R equals 0.072 Meters, which is less than 3 inches.
(Please don't hold any errors against me. Check it yourself!)

Since this distance is well within the near-field region, there is an
uncertainty that depends on whether there is electric coupling or magnetic
coupling. To be absolutely sure there is no interference, you could back off a
couple of wavelengths, say fifteen feet on two meters. (Others may say ten
times 3 inches would be enough.)

How does this one Volt per Meter assumption compare with the defibrillator
specifications? If the spec is higher, then there seems little to worry about.

73, Fred, K4DII