Re: Toroid capacitance reduction

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

Original poster: "Dr. Resonance" <resonance-at-jvlnet-dot-com> 


We measure the toroid cap stack on cardboard boxes at the exact same height
as the toroid would be operating atop the coil.  It always measures 10-20%
lower atop the coil.  I've always attributed this to the metal copper turns
of the coil acting as a shield for the electrostatic lines eminating from
the lower section of the toroid.

With the effect of shielding turns on the coilform the toroid's capacitance
value should always be less not more because the toroid is the exact same
dimensions from all other points in the room.  The coil's turns are acting
as a shield to cut off the electrostatic lines.

I've measasured this with 12, 20, 24, 30, 34.5, and 48 inch major dia.
toroids.  It seems to be consistent when toroid dimensions are equal to 2
times the coilform dia.

Dr. Resonance

Resonance Research Corporation
E11870 Shadylane Rd.
Baraboo   WI   53913
 >
 >  > I do not believe anyone has ever explained this Toroid reduction
 >  > with Tesla coils. It should be an increase because the toroid is
 >  > closer to other objects. Maybe Paul or Antonio will look into it
 >
 > John's comments about reduction of toroid capacitance when it is
 > in-situ over the coil are quite correct.  However, there is no
 > mystery to the reason why - it is a well known and understood
 > phenomena in electrostatics.
 >
 > Generally if you have two objects A and B, having self capacitances
 > of Ca and Cb respectively when each is measured (or calculated) in
 > isolation, then when they are brought together, joined or just close
 > to one another, the combined capacitance of the two will always be
 > less than Ca + Cb.
 >
 > You can think of it as each object shielding the other to some
 > extent.  The effective electrical 'surface area' of the two objects
 > combined is less than the sum of the two separate objects.  This is
 > because charge which would have been distributed fairly evenly over
 > each of the separate objects will be displaced away from the other
 > object when the two are brought together.
 >
 > With fixed width font and some allowance for the ascii artwork,
 > the charge distribution around two isolated blobs is something
 > like:
 >
 >       +  + +                      + + +
 >     +  ____  +                  + ____  +
 >    +  /    \  +               +  /    \  +
 >   +  |      |  +             +  |      |  +
 >    +  \____/  +               +  \____/  +
 >     +        +                  +       +
 >       + + +                       + + +
 >
 > where the + indicate charges on the surface of each blob (not in
 > the air around it as the sketch might suggest!)
 >
 > When they are brought close to each other, we get
 >
 >            +  + +   + + +
 >          +  ____  +  ____  +
 >         +  /    \   /    \  +
 >        +  |      | |      |  +
 >         +  \____/   \____/  +
 >          +        +        +
 >            + + +     + + +
 >
 > If you count the charges (ie measure the capacitance) you see
 > there are now fewer, because charge is not inclined to occupy the
 > surfaces of the objects which are close to each other.  Thus the
 > total charge needed to raise the combined objects to some given
 > potential is less than the sum of the amount needed for the two
 > separate objects.
 >
 > Why doesn't charge want to sit on the adjacent surfaces?  Because
 > each charge can find a lower potential energy by moving elsewhere.
 >
 > Each charge is immersed in the E-field generated by all the other
 > charges, and those that are free to move (ie not bound into atoms,
 > etc) will 'fall' through the field until they find their lowest
 > level.  In effect the charges are just trying to get as far away
 > from each other as they can.  The upshot is that less total charge
 > is needed to reach some potential because the adjacent surfaces
 > don't need to be covered.
 >
 > It isn't too hard to calculate this behaviour, because at low
 > frequencies the only significant force on each charge is just the
 > Coulomb force.
 >
 > Programs made available by Terry, Bart, and more recently Antonio,
 > will all take these effects into account quite accurately, because
 > one way or another they compute the actual charge distribution of
 > the in-situ objects.
 >
 > And as a corollary, not only does C_toroid fall when in-situ, so
 > does C_secondary too.
 >
 > Apparently my hands are waving so much they risk bringing down the
 > xmas decorations, but hope that helps!
 >
 > Greetings All,
 > --
 > Paul Nicholson
 > --
 >
 >
 >