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Forces between charged particles

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109.


    What would be the force between two 1.7-mm-diameter copper spheres 1.7cm apart? Assume that each copper atom has an equal number of electrons and protons.
    Express your answer using two significant figures.

    2. Relevant equations
    M = ρ V
    ρcopper = 8920 kg/m^3
    Vsphere = 4*pi / 3 * r^3
    Molar mass of Copper = 63.5 g/mol
    1 mol = 6.02*10^23
    upload_2014-9-27_15-39-42.png
    e=1.6*10^-19 C
    k=9*10^9 N*m^2 / C^2
    total negative charge = -e * #electrons
    total positive charge = e * #protons

    3. The attempt at a solution
    Msphere = (8920 kg/m^3)(4pi/3)(1.7/2 * 10^-3)^3 *1000g/1kg= 0.0229 g

    0.0229 g Copper * 1/63.5 mol/g * 6.02*10^23 atoms/mol = 2.17 *10^20 atoms of Copper

    total negative charge = -e*(29)*(2.17*10^20)
    total positive charge = (29)*(-e+10^-9)
    net charge = the sum of the two above

    The charge is the same for both copper spheres, so just plug and chug after this point. I am getting the wrong answer. I am certain it has to due with the calculation of the total charge on each of the copper spheres.

    I am not sure how to tie the "Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 10^9." part into the actual charge of the copper atoms.
    Please help--I dont understand what they mean.
     
    Last edited by a moderator: Sep 27, 2014
  2. jcsd
  3. Sep 27, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    Saying that something differs by one part in x, where x is some number, means that one of them is larger (or smaller) by an amount that is 1/x of the other.

    So in this case you might decide to let the proton charge be smaller than the electron charge by one part in 109, so then
    $$q_p = \left(1 - \frac{1}{10^9} \right) q_e$$
    Of course you could make the proton charge the larger one, too, if you wish.

    You could also just work with the differences in the charges, since ##\Delta Q = (q_e - q_p) = \frac{1}{10^9}q_e##
     
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