If it were conducting then it would not be required to be constantly driven over the pulley; the positive charge gained at the bottom could just flow directly to the large sphere.

  • $\begingroup$ The body of your question sounds like you are trying to post and answer to the title of your own question... Try to write and format your questions in a way that they are easy for other users to read and understand. Put some effort into it. You will get better answers. $\endgroup$
    – hft
    May 9, 2015 at 1:13

1 Answer 1


If the belt is not insulating, any charge on the terminal will just flow back to ground, so you can't build up charge on the terminal and it will not rise in potential. The point of a Van de Graaff generator is to physically move charge against the electrical gradient, and you can't do that if the belt lets it slip away.

Now, you can instead use the Pelletron system from National Electrostatics instead of the original High Voltage Engineering (the company Van de Graaff founded to make accelerators) rubberized belt. The Pelletron system uses chains of 'pellets', where each pellet is a plastic cylinder plated with metal on the circumference. The links are also plastic, so the chain is insulating along them. These suffer less degradation of the charge carrying capacity of the chain. Belts would wear and develop bald spots that wouldn't carry charge, so the terminal voltage would not stay constant.

  • $\begingroup$ In that case the ground the terminal will just acquire the same potential, and the belt will have no net charge on it, is that right? (So now more charge flows to the terminal.) $\endgroup$
    – Charles
    Apr 20, 2015 at 14:46
  • $\begingroup$ I believe that you are still fuzzy on how a Van de Graaff generator works. It mechanically moves charge from low potential to high potential, effectively across a capacitive gap, to generate a voltage that is proportional to the charge moved. If the terminal acquires the same potential as ground, well, it isn't at potential any more. There are purely electrical ways of generating high voltages, such as a Cockroft-Walton style supply. $\endgroup$
    – Jon Custer
    Apr 20, 2015 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.