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Homework Help: Coulomb's Law

  1. Aug 27, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    I've tried working on this for two days and can't figure it out.

    In (a), the cumulative force on A is the sum of the force from B and the force from C. or:

    [tex]F=\frac{k Q_a Q_b}{d^2} + \frac{k Q_a Q_c}{d^2}[/tex]

    In (b), the same applies.

    however, I can't figure out what to do with these equations in order to isolate Qb or Qc. If I use a negative d (-d) as a distance from A-->B in (b), then I get two equations that are identical, but shouldn't be. If I move the origin, it doesn't seem to matter either.

    I don't know how to start this problem.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 27, 2009 #2
    Qc/Qb=1.328 (sorry)

    Ok, so in the first scenario, both charges B and C exert forces to the left, on charge A, since all charges are positive. However, in figure b, charge b exerts a force to the right, while charge C exerts a force to the left, on charge a. Hence, in fig. a,
    while in fig b

    If you factor out k, QA, and r^2, keeping in mind that RB=RC (distance from a to b, and a to c are the same in both figures), and divide the two equations, you can get QC/QB which is 1.328

    Hope this helps,
    Last edited: Aug 27, 2009
  4. Aug 27, 2009 #3
    Because of the direction of the force you know that Qc is bigger than Qb so write an equation like you did for part a but getting the directions right.Now divide one equation by the other and tidy it up.
  5. Aug 27, 2009 #4
    You are correct, QC should be larger than Qb, but when I worked it out, I got QC/QB==1.328, I don't know why...

    corrected, see above, I had a problem with the signs in my initial equation...
    Last edited: Aug 27, 2009
  6. Aug 27, 2009 #5
    I'm still absolutely lost on this.

    Faraday, I understand what you did taking into account the negative Force due to B in the second example. However, I don't know what to do with the equations at this point.

    I end up with:

    (a) F=(kQa/d^2)(Qb+Qc)

    (b) F=(kQa/d^2)(Qc-Qb)

    I don't understand where to go from here.
  7. Aug 27, 2009 #6
    You can now divide the two equations above, so F1/F2==(QB+QC)/(QC-QB), hence, qc/qb is 1.328. (I had a problem with my signs in the initial solution.)
    Last edited: Aug 27, 2009
  8. Aug 27, 2009 #7
    Why should I do that? I am not following the logic, I guess.
  9. Aug 27, 2009 #8


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    Homework Helper

    Both equations have "d", which you don't know but can eliminate by dividing the equations. After dividing, you have the ratio F1/F2, which you can calculate, as well as Qb and Qc. You'll have to rearrange to get an expression for the ratio Qb/Qc.
  10. Aug 27, 2009 #9
    Okay. 1.328 is right, and I did the simplification on paper here as well. Ends up looking like:

    [tex]\frac{Q_c}{Q_b} = \frac {F_1+F_2}{F_1-F_2}[/tex]

    I don't know if I ever would have figured out to divide one by the other though.
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