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Compressional force

  1. Sep 1, 2010 #1


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    1. The problem statement, all variables and given/known data

    Suppose 7.00 g of hydrogen is separated into electrons and protons. Suppose also the protons are placed at the earth's North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth?

    2. Relevant equations

    Fe = Ke*|q1||q2|/r^2, radius of earth is 6.38*10^6 * 2 to get diameter is 1.3*10^7...

    3. The attempt at a solution
    I believe this is the correct way:

    7 * 6.02^1023 = 4.21*10^24 (since it is 7 hydrogens, and 1 hydrogen = 1 Avagrado's number)
    = [(4.21*10^24)*(1.60*10^-19)]^2/(4^(6.38*10^6)^2) * 8.99*10^9 = 2.51*10^7. I think the reason we multiply by 4 is because we need to get the diameter for both proton and electron, giving 4.
    Last edited: Sep 1, 2010
  2. jcsd
  3. Sep 2, 2010 #2

    Andrew Mason

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    I don't follow your reasoning.

    In order to determine the force between 7 moles of protons and 7 moles of electrons at opposite poles of the earth you have to know the permittivity of the earth. Since the earth is largely iron, look up the permittivity of iron.


    [tex]F = \frac{Q_1Q_2}{4\pi\epsilon r^2}[/tex]

    where [itex]\epsilon = \epsilon_r\epsilon_0[/itex] ([itex]\epsilon_r[/itex] being the relative permittivity of iron).

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