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Another question on electric fields.

  1. Mar 29, 2008 #1
    Another question on electric fields. :(

    I have another problem I don't understand on electric fields. I thought I understood these after my last post, but I guess not. I'm guessing it has something to do with the signs of the forces, but I am not sure.

    1. The problem statement, all variables and given/known data
    Particles of charge Q1 = +68 µC, Q2 = +49 µC, and Q3 = -80 µC are placed in a line (Fig. 16-37). The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two.

    [​IMG]
    Figure 16-37


    2. Relevant equations
    F = KQ_1Q_2/r^2


    3. The attempt at a solution

    F = K * Q1Q2/r^2
    F_12 = (K)(6.8 X 10-5)(4.9 X 10-5)/.35^2
    F_12 = 244.8 N (goes to the left, repels)

    F_13 = K(Q1Q2)/r^2
    F_13 = (K)(6.8 X 10-5)(4.9 X 10-5)/.7^2
    F_13 ~= 100 N (goes to the right, attracts)

    F_q1 = -244.8 + 100
    F_q1 = -144.8 N (wrong)

    However, the second part of the question asked which direction it went in, and based on my answer, I put "left" and that was correct. I'm confused.
     
  2. jcsd
  3. Mar 29, 2008 #2
    What does F_13 mean? 1 on 3 or 1 due to 3?
     
  4. Mar 29, 2008 #3

    nrqed

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    Watch out! In F_13 you used the charge of q2 instead of q3!
     
  5. Mar 29, 2008 #4

    Doc Al

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    Your answers look good to me. Since they ask for the direction separately, I suspect that for the first part they just wanted the magnitude of the force.
     
  6. Mar 29, 2008 #5

    Doc Al

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    That was a typo; the answer matches the correct numbers. :wink:
     
  7. Mar 29, 2008 #6

    nrqed

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    Ah, ok! :smile: I should have checked!

    To the OP: yes, they surely want the magnitude only. And be careful with sig figs.
     
  8. Mar 29, 2008 #7
    Wow. That was dumb of me for not seeing magnitude right next to the problem...thank you everyone in this thread. At least I know what I'm doing now...
     
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