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![]() "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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