Michel Jullian wrote. > > Fred I agree with your formula for like charges, but only if the totality of the Earth's charge is kind enough to come right under your VDG, which may be difficult since it will be repelled by it. > > What is more likely IMHO is that your VDG's charge, much lower but much more concentrated than the Earth's, will repel not the Earth but it's dilute like charge some distance away, attracting instead opposite charge right under it (this may be what is called "image charge"?). > That argument holds for free (electron) charge as in a metal, but not for electrons attached to atoms-molecules in the earth (soil or `plants, buildings, etc, where the Electron Affinity ranges from a fraction of eV to 2.5 eV or more.
You can see the effect from this in the Crop Circles in your breakfast cereal or fuel ethanol fields out in the country. :-) > > This would result in an increase of the VDG's apparent weight as you have observed, irrespective of the sign of the VDG's charge, or of whether it has been drawn from the Earth or not (the Earth being such a large charge reservoir comparatively). > That remains to be determined. > > Keep us tuned, we'll find a way to fly you to the moon somehow ;-) > My present problem is determining the field configuration between a sphere and two (or more) oppositely charged spheres connected by tubular passageways and the resulting force on terra firma at a few meters altitude. :-) Fred > Michel > > ----- Original Message ----- > From: "Frederick Sparber" <[EMAIL PROTECTED]> > To: "vortex-l" <[email protected]> > Sent: Monday, January 29, 2007 7:56 PM > Subject: Re: [Vo]: Re: Re Van de Graaf Antics > > > > Why I think an isolated-charged conductor (Sphere or Cylinder) > > partially surrounded by a conductor (Sphere or Cylinder) > > of opposite charge will repel the earth's charge. > > > > http://f3wm.free.fr/sciences/jefimenko.html > > > > In spite of your skepticism Michel: > > > > F = k*q*Q/height^2 newtons. :-) > > > > Fred >
