Re: Magnifier topload size?

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

Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <teslalist-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
 >
 > Original poster: "Jeff W. Parisse by way of Terry Fritz 
<teslalist-at-qwest-dot-net>" <jparisse-at-teslacoil-dot-com>

 > My thinking (I've been wrong before) is that the above is true only at
 > high voltage (true, that's where it really counts) but wouldn't the
 > theoretical capacitance (read: low voltage) be based on the surface area
 > of the plate/dielectric. Given Faraday's law (charge on the outside) we
 > can only consider the portion of Electrum's rings facing outward (I
 > reason). Therefore the ringed sphere would have about half the surface
 > area (capacitance) than the ideal sphere?

Remember the discussions about a thin toroid having great part of the
capacitance of a sphere with the same radius? A wire sphere is a
better approximation. The difference is in breakdown voltage, that
is certainly smaller for a wire sphere, because almost the same charge
that would be distributed over the surface of the sphere will be
distributed over the outer surfaces of the wires, concentrating
the electric field there.

 > Now let's turn it on...
 >
 > Now, we know that there is a "virtual surface" of electrons that builds
 > up before breakout that make's our toroid "grow". Add I believe I
 > understand you to say that the "growth" between rings make the surface
 > "electrically continuous".

Then, a wire sphere at high voltage behaves as a somewhat bigger
sphere, and has more capacitance than a sphere of the same size.

Antonio Carlos M. de Queiroz