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Coulomb's Law and repulsive force

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Charges q1 and q2 exert repulsive forces of 15 N on each other. What is the repulsive force when their separation is increased so that their final separation is 160% of their initial separation?

    a.1.3 N

    b.9.8 N

    c.8.7 N

    d.5.9 N


    2. Relevant equations

    F = (k*q_1*q_2)/(r^2)

    3. The attempt at a solution

    F_2 = [k*q_1*q_2]/[(1.6^2)*(r^2)]

    F_2 = F/(1.6^2) = F/2.56

    F_2 = 15 N/2.56 = 5.86 N ??




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

    The force between two very small charged bodies is found to be F. If the distance between them is doubled without altering their charges, the force between them becomes


    a.F/2

    b.2F

    c.F/4

    d.4F


    2. Relevant equations

    F = (k*q_1*q_2)/(r^2)

    3. The attempt at a solution

    F_2 = (k*q_1*q_2)/(2r)^2 = (k*q_1*q_2)/(4*r^2)

    F_2 = F/4 ?


    Thanks.
     
  2. jcsd
  3. Jan 21, 2007 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah I think you've nailed the inverse square concept.
     
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