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Magnitude of total electrostatic force on a third particle ?

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle with charge −8 μC is located on the x-axis at the point 8 cm, and a second particle with charge −6 μC is placed on the x-axis at −6 cm.

    What is the magnitude of the total electrostatic force on a third particle with charge −4 μC placed on the x-axis at −2 cm?
    Answer in units of N

    2. Relevant equations

    The Coulomb constant is 9.0× 10^9 N·m2/C2.
    Fe = (kq1q2/r^2)

    3. The attempt at a solution

    [(9*10^9)(4*10^-6)(8*10^-6)]/(0.1)^2 + [(9*10^9)(4*10^-6)(6*10^-6)]/(0.04)^2

    The final answer I get is 163.8 N, which is incorrect.
     
  2. jcsd
  3. Mar 2, 2012 #2

    gneill

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    Staff: Mentor

    Check the directions in which the forces act.
     
  4. Mar 3, 2012 #3
    I don't understand. Does that mean I subtract the two forces?
     
  5. Mar 3, 2012 #4

    gneill

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    Staff: Mentor

    Forces are vectors. They add as vectors. So after you've determined the magnitudes of the forces you need to determine the directions they're acting, too. "Add" them accordingly. Drawing a sketch of the scenario will help.
     
  6. Mar 3, 2012 #5
    That's exactly like I did.
    Apparently it was wrong.
    Please take a look at my work.
     
  7. Mar 3, 2012 #6
    Okay, so all of the charges are negative. So between the middle (main) charge and the left charge, it wants to move right. Between the middle and the charge to the right of it, it wants to move left. The magnitude of the electrostatic force of the middle charge and the left charge is 135N. The magnitude of the electrostatic force of the middle charge and the right charge is 28.8N. Since the charge wants to go different directions in those situations, shall I subtract the two forces yielding an answer of 106.2N?
     
  8. Mar 3, 2012 #7
    Correct.
     
  9. Mar 3, 2012 #8

    gneill

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    Staff: Mentor

    I did look at your work, and one of the forces is acting in the wrong direction.

    Try this. Write the calculation for each force individually and state its magnitude and direction.

    EDIT: Never mind, I see you've done essentially that above. Good show :smile:
     
  10. Mar 3, 2012 #9
    Thanks guys!
     
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