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Coulomb's Law and silver spheres

  1. Feb 6, 2006 #1
    Two small silver spheres, each with a mass of 12.0 g, are separated by 1.00 m. Calculate the fraction of the electrons in one sphere that must be transferred to the other to produce an attractive force of 3.00e4 N (about 3 tons) between the spheres. (The number of electrons per atom of silver is 47, and the number of atoms per gram is Avogadro's number divided by the molar mass of silver, 107.87 g/mol.)

    Any help would be appreciated....i'm so confused with this material right now:frown:

    I'm guessing you will use coulombs law of course (F=k[q1q2/r^2])
  2. jcsd
  3. Feb 6, 2006 #2


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    With the information at hand, you can calculate how many electrons there actually are in each sphere. Also, as it is, both sphere are neutral: in each self respecting silver atoms, there are as many protons as there are electrons, making the total charge on each sphere 0 C. And you also know the distance btw the spheres:1.0m. So, using Coulomb law, find what charge magnitude q is necessary to create an attrative force of 3.00e4 N btw two charges a separated by 1.0m. To how many electrons does that charge correspond? That is the amount of electrons you must transfer from one sphere to the other. Finally, what fraction of the total number of electron does that represent?
    Last edited: Feb 6, 2006
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