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Homework Help: Coulomb's Law - Calculating Net Force

  1. Jan 24, 2015 #1
    Hello all,

    I am new to the website and I have joined because I just started a Physics class in College and I've never done any sort of Physics and after my first day I was thrown this problem and have been looking at it for a while now and finally gave in for some sort of help.

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

    Particles of charge +70, +48, and -80 μC are placed in a line. The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two.

    2. The attempt at a solution

    Before posting, I looked to see if any other posts were similar and I had came across one from 2005 and I've looked at his math:
    I understand the user, neonerd, had said that he had made a mistake with the -9 on (8.99*10^-9) however I am not making the connection between

    "F = (8.99*10^-9) (48*10^-6) (70*10^-6) / 0.35^2
    = 2.46*10^-16"

    I understand the formula is F = K q1 q2 / r^2

    My professor has k=9x10^9 Nm^2/Coul^2 instead of using k=8.99x10^9 but says that meters and C cancel out so I know why K is different of neonerd's work

    So the formula would start out as
    F=(8.99x10^9) q1 q2 / r^2.
    R is the distance so 0.35m^2
    F=(8.99x10^9) q1 q2 / 0.35m^2
    I am not understanding why the q1 and q2 are (70*10^-6) (-80*10^-6) and how neonerd got (2.46x10^2N)

    Again I apologize if my misunderstanding is hair-pulling to anyone I just have never seen this and I would appreciate any help.
  2. jcsd
  3. Jan 24, 2015 #2


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    9 is just a rounded version of 8.99, and the minus in the exponent was a mistake.
    with Coulomb there. The charges are defined to have those values in the problem statement (µC, "micro-coulomb", means 10-6 Coulomb).
    If you want to calculate the force between the 70µC and the -80µC charge: the distance between the charges is not 0.35m there.
  4. Jan 24, 2015 #3


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    They only cancel out if all the distances plugged in are in metres and all the charges plugged in are in Coulombs.
    That's not strictly correct. Since F, q1, q2 and r are represented symbolically at this point, their units are unspecified in the equation. Therefore the units of the constant should be shown:
    F=(9x109) q1 q2 / r2 Nm2/C2
    When the values are plugged in:
    F=(9x109) (70μC)*(48μC) / r2 (Nm2/C2)
    =(9x109) 70*(48) / r2 (Nm2/C2) μC2
    = (9x109) 70*(48) / r2 Nm2 (μC/C)2
    = (9x109) 70*(48) / r2 Nm2 (10-6)2
    = (9x10-3) 70*(48) / r2 Nm2
    = (9x10-3) 70*(48) / (0.35m)2 Nm2
    = (9x10-3) 70*(48) / (0.35)2 N
    neonerd got that number for the case where the charges are 48 and 70 μC.
  5. Jan 24, 2015 #4
    Micro! I feel very slow for not realizing it was micro That explains why it was 10^-6. Thank you! Also, I would have to use 0.7m^2 correct if it was from q1 to q3?
    I am trying the math and my numbers are not adding up. I am usually great with math but -- well I'll just show my work.

    I see! I had just crossed out mine but I think practicing it the way you're showing would be better for the future because I am sure it will come up and try to haunt me. I'll be ready for it though!

    F= K q1 q2 / r^2
    =(9*10^9N*m^2/C^2) (70*10^-6C) (48x10^-6C) / 0.32^2 (0.35^2 Because it is from q1 to q2)
    = I ended up canceling out the C^2 and m^2 even though I understand haruspex's work and ended up with:
    (9x10^-3N) (70) (48) / 0.35^2 which is similar to haruspex's work but I left the N in the K area.
    =Multiplying the top I got 30.24N/0.35^2
    =Finally I get 246.857

    I thought that it was my signs and double check and triple checked but I am not getting -1.4x10^2 or any numbers around 140.


    Okay I am seeing why. I am needing to come up with k q1 q3/r^2 THEN add them (I think)


    Yes! After doing the equation for q1 q3 I added up the two and got the answer! Thanks mfb and haruspex for your patience and time! I appreciate it now to do the other two parts of this problem! :D
    Last edited: Jan 24, 2015
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