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Two nonconducting spheres problem

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Two nonconducting spheres have a total charge of 13.4 μC. When placed 0.362 m apart. the force of repulsion is 2.777 N. What is the value of the greater charge (Give your answer in coulombs)?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 7, 2008 #2

    Doc Al

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    Staff: Mentor

    Hint: Coulomb's law.
     
  4. Mar 8, 2008 #3
    I tried this equation below because I just put in the total charge for Q and got the wrong answer. I dont understand how I get answers for two different charge swhen I work through it.
    E= k (q1+q2) / r^2
     
  5. Mar 8, 2008 #4
    well first equation is F = K(q1)(q2)/r^2 and second equation is q1q2=13.4 μC
    2 equations 2 unknowns.
     
  6. Mar 8, 2008 #5

    Doc Al

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    Staff: Mentor

    You are correct, but that second equation has a typo: it should be q1 + q2 = 13.4 μC
     
  7. Mar 9, 2008 #6
    F=kq1q2 / r^2 q1 + q2 = 13.4x10^-6C
    2.777= (8.99x10^9)q1q2 /(0.362)^2
    q1q2=4.05x10^-11
    q1=(4.05x10^-11)/ q2

    (4.05x10^-11)/q2 + q2 = 13.4x10^-6C
    (4.05x10^-11)/ q2 + (q2^2)/q2 = 13.4x10^-6C
    (4.05x10^-11) + (q2^2) = (13.4x10^-6C)q2
    quadratic, when solved q2=4.60x10^-6, q2= 8.80x10^-6

    Both of these numbers when added up will equal the total charge and when I sub them into the equation q1=(4.05x10^-11)/q2 I get the same thing. However this is not the correct answer, could someone point out where I went wrong?
     
    Last edited: Mar 9, 2008
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