You've made me think about it a lot more now. The AC/DC resistance argument
is certainly valid. It does indeed mean that any capacitive field will be
captured with less loss than letting it end up in lossy ground, or it would
if the sheet covered the area captured by the E field due to the WHOLE
However, just concentrating on the loading coil, which has a huge H field
and very low E component is a different matter. What is the loss due to
the H travelling through ground of poor conductivity ? - not too much I
More thought required..............
I thought it means that the current density over the thickness of the foil
is almost uniform with result that AC and DC resistance are almost equal.
But even with the thin foil that resistance will be so much smaller than
resistance via the real earth that I expect almost all capacitive current
from the coil is captured by the foil.
Am I wrong?
73, Dick, PA0SE