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Potential Difference in a van der graaff generator

  1. Oct 4, 2011 #1
    Hi, this problem is from 'Electromagnetism' by Grant & Phillips and it states,

    1. The problem statement, all variables and given/known data
    'http://i1129.photobucket.com/albums/m505/physicsbird1/126fig.png" [Broken] shows a cross-section of the cylindrical high-voltage terminal of a van der Graaff generator, surrounded by an 'intershield' and a pressure vessel, both of which are also cylindrical. The gas in the pressure vessel breaks down in electric fields greater than 1.6 x 10[itex]^{7}[/itex] volts/m. If the radii of the terminal, intershield and pressure vessel are 1.5 m, 2.5 m and 4 m respectively, what is the highest potential difference that can be maintained between the terminal and the pressure vessel?'

    The answers given in the back of the book: '31.1 MV. When the terminal is at this voltage the intershield must be at 18.8MV to prevent breakdown.'


    2. Relevant equations
    d∅ = -E[itex]\bullet[/itex]dr

    E = [itex]q/4πε₀r^{2}[/itex]

    3. The attempt at a solution

    1.6x10[itex]^{7}[/itex] = [itex]q/4πε₀(2.5)^{2}[/itex]

    1.00x10[itex]^{8}[/itex] = [itex]q/4πε₀[/itex] (Let [itex]q/4πε₀[/itex] = A)

    A = 1.00x10[itex]^{8}[/itex]


    ∅[itex]_{B}[/itex] - ∅[itex]_{A}[/itex] = -1.00x10[itex]^{8}[/itex][itex]\int^{4}_{1.5}[/itex](1[itex]/r^{2})[/itex]

    = -1.00x10[itex]^{8}[-1/r]^{4}_{1.5}

    = -1.00x10^{8} [-1/4 + 1/1.5]

    = -4.17x10^{7}

    So... can anybody tell me where I've went wrong?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 5, 2011 #2
  4. Oct 5, 2011 #3
    You wrote, "E = q/4πε₀r^2"

    E goes as 1/r ?
     
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