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"?). 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). Keep us tuned, we'll find a way to fly you to the moon somehow ;-) Michel ----- Original Message ----- From: "Frederick Sparber" <[EMAIL PROTECTED]> To: "vortex-l" <vortex-l@eskimo.com> 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
