Re: determining capacitance of a topload?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:
 >
 > Original poster: "tmoorezz" <tmoorezz-at-adelphia-dot-net>
 >
 > Hello everyone,
 >
 >      Just a quick question, I found the surface area of an irregularly 
shaped
 > toroid which is 722 in squared, what do I have to do now to figure out the
 > capacitance, Thanks!

It doesn't have much relation to the surface area, because the charge
distribution on the surface is not uniform (almost no charge in the
"hole").
See:
http://www.coe.ufrj.br/~acmq/tesla/capcalc.pdf
For a good approximation, use:
If d/dd<0.25: C=100e-12*1.8*(dd-d)/Ln(8*(dd-d)/d)
else C=100e-12*(0.37*dd+0.23*d),
where d is the minor diameter and dd is the major diameter. In meters.
The result is in Farads.
Ex: a 50 cm x 10 cm toroid:
The formula gives 20.77 pF
The exact calculation results in 20.74 pF
This is the free-space capacitance. The presence of the coil below the
toroid reduces the effective value a bit.

Antonio Carlos M. de Queiroz