Confusion: Van de Graaff generator

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SUMMARY

The discussion centers on the operational principles of a van de Graaff generator, specifically addressing the movement of charge between a small conducting sphere and a larger outer shell. The physics book explains that charge q moves to the outer surface of the shell due to the nature of conductors. However, a participant raises concerns about the implications of creating a hole in the outer shell, questioning the application of Gauss' Law and the uniform potential of the outer shell. The confusion arises from the interaction of charge flow and the physical structure of the generator.

PREREQUISITES
  • Understanding of electrostatics and charge distribution
  • Familiarity with Gauss' Law and its applications
  • Knowledge of conducting materials and their properties
  • Basic principles of electric potential and field theory
NEXT STEPS
  • Research the principles of charge movement in conductors
  • Study Gauss' Law and its limitations in real-world applications
  • Explore the design and functionality of van de Graaff generators
  • Investigate the effects of small openings in conductive shells on electric fields
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Physics students, electrical engineers, and anyone interested in the principles of electrostatics and the operation of particle accelerators.

conquerer7
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My physics book explains a van de Graaff in this way:

A small conducting sphere of radius a and carrying charge q is located inside a larger shell of radius b that carries charge Q. A conducting path is momentarily established between the two conductors, and the charge q then moves entirely to the outer conductor, because the charge on a conductor always moves to its outer surface.

But in any actual machine, you'd have to punch a hole in the outer sphere to bring the charge inside; wouldn't that merge the inner and outer surfaces? You can't even apply Gauss' Law like usual because the Gaussian surface would have to pass through the hole, and there'd be a nonzero field there.

Am I just completely misunderstanding it?

Edit: In addition, since the potential of the outer shell is so big (and should be uniform throughout), why doesn't charge flow off it onto the belt?
 
Last edited:
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The hole is small enough that its influence is negligible.
 

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