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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

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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.

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  • $\begingroup$ 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 '14 at 22:00
  • $\begingroup$ The yellow one - or more precisely - half of the yellow line. $\endgroup$ – SpiderPig Jan 26 '14 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 '14 at 7:21
  • $\begingroup$ Yes it doesn't matter whether the sphere is hollow or not. $\endgroup$ – Sandesh Kalantre Jan 27 '14 at 7:39

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