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Old August 24th 08, 10:38 PM posted to rec.radio.amateur.antenna
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Default E-Field between 2 parallel wires


"Jon Mcleod" wrote in message
m...
MAYBE NOT EXACTLY THE RIGHT GROUP, but..

If I tape 2 insulated, parallel wires to the wall, x cm apart, and then
drive a sinewave into them (Vo p-p), how can I calculate the field
strength between the 2 wires?

For instance, 120KHz, 100V, 5cm apart, what is the field in V/cm between
the two wires?

Any references or information on how to calculate this would be greatly
appreciated.


if you are looking for a closed form solution you will need to investigate
some classical electrodynamics methods for wires on a dielectric half plane.
There are probably some not so elegant methods to calculate it using the
relative dielectric constants of the air on one side and the wall material
on the other. The field will of course be different in the wall material
than in the air. as a first approximation you can take an average of them
and use that to estimate the field as if the conductors were embedded in an
infinite block of dielectric material, that is probably a simpler problem.

there are programs that will do finite element analysis and come up with
numerical solutions, that is probably simpler in most cases than the raw
field equations.

For a relatively low frequency like 120khz and those low voltages you can
bound the problem by assuming 2 wires in air... 100v/5cm=20v/cm half way
between the wires... with a material on one side it will usually have a
lower field than in the air by approximately the ratio of the relative
dielectric constants. For wood the constant depends a lot on moisture
content, from a low of maybe 5 up to 100 for saturated wood. The higher the
dielectric constant the lower the field in the material, as a first
approximation with a constant of 5 you could guess that 80% of the field was
in the air and 20% in the wall... and of course the farther from the wires
the lower the field... so from a max of maybe 20v/cm between the wires the
field will decrease as you move away from the wall in air, and decrease
about 5x faster as you move into the wall. the hard part is close to the
wires, there it depends on the diameter of the wire, the smaller the
diameter the higher the field, to get those numbers you need one of the
other methods above.


 
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