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Question about Van De Graaf generator

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    I have a sentence in my textbook that reads:

    Because the "breakdown" electric field in air is about 3 X 106 V/m, a sphere 1.00m in radius can be raised to a maximum potential of 3 X 106 V

    And now the problem states: The metallic sphere on top of a large van de Graaf generator has a radius of 2.0 m. Suppose that the sphere carries a charge of 3x10-5 C


    2. Relevant equations
    V = Q/4∏rε0


    3. The attempt at a solution
    Does that mean to find the Volt, I have to simply multiply that 3 X 106 V by 2, to give me 6 X 106 V?

    Or do I have to use the above equation to find the volt inside the generator?
     
  2. jcsd
  3. Oct 2, 2012 #2

    gneill

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    Staff: Mentor

    Why not employ your Relevant Equation?
     
  4. Oct 2, 2012 #3
    I guess it's the line from the text that confused me as to whether that equation was even necessary or not
     
  5. Oct 2, 2012 #4

    gneill

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    Staff: Mentor

    The line in the text is telling you what limits the maximum potential on a surface (before breakdown and arcing occur to drain off the charge). If your calculated potential approaches this value, stand well back from the apparatus!!!:eek:
     
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