Only asking those questions because I can't afford to get a real generator.

  1. If the voltage on the sphere was 50 kV, then I used a metal rod to discharge it and I connected a voltmeter to wire grounding the rod, will it read 50 kV?

  2. Does the formula $V=kq/r$ apply for the generator? (According to some sites, it does.) If it does, which radius is meant? The hollow part radius? Will the thickness of the wall matter? enter image description here


1 Answer 1


If the voltmeter is able to measure such high voltages and has a very high internal resistance you can connect it between the sphere and ground and it will display 50kV.

The thickness of the wall doesn't matter at all. So r is the external radius.

  • $\begingroup$ so according to the photo i just uploaded, which one is the r in the formula , the red one or the yellow one? $\endgroup$
    – user28324
    Jan 26, 2014 at 22:00
  • $\begingroup$ The yellow one - or more precisely - half of the yellow line. $\endgroup$
    – SpiderPig
    Jan 26, 2014 at 22:08
  • $\begingroup$ i asked a friend he said : the electrostatic excess charge will be on the surface of the sphere , so it won't matter if the sphere was hollow or not, is this statement right? $\endgroup$
    – user28324
    Jan 27, 2014 at 7:21
  • $\begingroup$ Yes it doesn't matter whether the sphere is hollow or not. $\endgroup$ Jan 27, 2014 at 7:39

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.