>         Secondly, lets suppose that there was enough wire in the cable to
create
>a significant capacitance. The effective circuit would look like a

RG-58 is about 30 pF/foot, according to my Belden catalog.


>capacitor with a resistance in parallel (we can ignore the inductance, it
>certainly won't make much difference beyond seconds of time). The value of
>the effective parallel resistance can be approximated by the resistance of
>the cable. Assuming the cable is copper and of a resonable diameter, lets

Sorry, if you take a hunk of coax with both ends open, any parallel
resistance _across_ the cap has to be from the insulation resistance of the
dielectric.

I don't have any teflon coax, but I'll bet it makes a _very_ good
low-leakage cap.

Look, at low frequencies, the capacitance of coax is like any other cap.
It stores a charge.  Put enough voltage across that cap and it can bite
you.  Use a high-quality dielectric and the stored charge will be there for
quite some time.


>for an extream example. The spool of wire holding 10 miles of wire would
>probably barely fit on my desk. So, I guess if you had a huge spool, you

I've got a spool that's about 26 miles long, and it isn't very big.  It's
not coax, and it's only 34 gauge.  Too bad I don't have access to both
ends--it'd be a neat inductor!


newell