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Potential due to point charges

  1. May 15, 2005 #1
    How much voltage must be used to accerlerate a proton (radius 1.2E-15 m) so that it has sufficient energy to just penetrate a silicon nucleus? A silicon nucleus has a charge of +14e, and its radius is about 3.6E-15 m. Assume the potential is that for point charges.
     
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  3. May 15, 2005 #2

    Andrew Mason

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    What is the potential energy of a charge q1 a distance r from another charge q2? What is the potential energy of a proton whose surface is situated 3.6E-15 m. from the centre of a silicon nucleus?

    AM
     
  4. May 15, 2005 #3
    that's all from the question given...
    I think it suppose r starts from infinity
    but i can't get to the model ans 4.2E-6 J
     
  5. May 15, 2005 #4

    OlderDan

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    You need to check the exponent in that answer, and you need to interpret "just penetrate" as the outer surface of the proton just barely penetrating the outer surface of the silicon nucleus.
     
  6. May 16, 2005 #5
    sorry, it should be 4.2E6 J
    I found the question quite confusing also....
    from the model ans and equation V = KQ/r
    r = 4.8E-15 m, which is the sum of proton and silicon nucleus radii
    but i don't know the reason behind....
     
  7. May 16, 2005 #6

    Andrew Mason

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    The force between charges depends on the distance from the centre of the charge. The electric field of a charge is given by Gauss' law, which states that the field depends only on the enclosed charge:

    [tex]\int_S E\cdot dA = \frac{q_{encl}}{\epsilon_0}[/tex]

    For a sphere:

    [tex]\int_S E\cdot dA = 4\pi r^2E = \frac{q_{encl}}{\epsilon_0}[/tex]

    [tex]E = \frac{q_{encl}}{4\pi \epsilon_0 r^2} = \frac{kq_{encl}}{r^2}[/tex]

    So if you assume that a proton and a silicon nucleus are uniform spheres, the field at their surface depends only upon the distance of that surface from the respective centres.

    AM
     
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