Scottf:
Sorry about the late reponse on this. I haven't been checking the forums as often as I used to...
The relation between capacitance and insulation does seem counterintuitive at first, but I think I can explain it in language that is more or less English.
Capacitance changes with respect to the electric field applied across the insulating dielectric. As the distance between the plates is increased, the region of space where the electric field is stored increases. As the distance increases, the electric field remains constant. This increases the potential difference between the plates, given by V = E*d where d is the distance between the plates. To keep the potential from increasing, charge must be removed from the plates. Hence, the capacitance decreases. Conversely, as you decrease the distance between the plates of the capacitor, the same electric field is distributed across a smaller distance, increasing the charge held on the plates.
You can also look at the following equation used to calculate capacitance for single conductor shielded cables, and note the relationship between capacitance and the thickness of the insulation...
C =(0.00736*K) / (LOG10(D/d))
(C is in units of microfahrads/1000ft in this case)
Where:
K = dielectric constant of the insulation
D = diameter over the insulation in inches
d = diameter of the conductor in inches (or over the conductor shield if present)
This relationship is used by some instrument manufacturing companies to measure insulation thickness if the dielectric constant of the insulation is known.
Dielectric losses for cables can be calculated by multiplying the capacitance by the power factor and the square of the operating voltage (and a conversion constant to get the losses in the units you want).
It's been a while since I had to talk in terms of electric fields, so my terminology may be a little rusty. Hope this helped.
Kraigb