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Magnitude of Force of two point charges on a third (electrostatics)

  1. Sep 4, 2013 #1
    Two point charges are placed on the x-axis as follows: charge q1 = 4.01nC is located at x= 0.201m , and charge q2 = 5.00nC is at x= -0.301m.

    What is the magnitude of the total force exerted by these two charges on a negative point charge q3 = -6.03nC that is placed at the origin?

    I know I need to use:
    F = (k * q_1 * q_2)/d^2, where k = 9E9


    I've tried taking the sum of magnitudes of F(1 on 3) and F(2 on 3), but have been incorrect. I set it up like below:

    F(1 on 3) = (k * 4.01E-9 * -6.03E-9) / 0.201^2 = -0.000005 N

    F(2 on 3) = (k * 5.00E-9 * -6.03E-9) / 0.301^2 = -0.000003 N


    This would give the answer as 0.000008 N, or more precisely 8.37E-6 N. Could anyone tell me where I'm going wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 4, 2013 #2

    rude man

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    check the sign of F(2 on 3). Which direction is it?
     
  4. Sep 4, 2013 #3
    Yeah, it should be positive, my bad. However I've also tried 0.000002 as an answer and that was incorrect:/
     
  5. Sep 4, 2013 #4

    gneill

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    Lose the fixed point notation and stick to scientific notation so that significant figures are clear. What's the value of the net force to three significant figures?
     
  6. Sep 4, 2013 #5
    F(1 on 3) = -5.387E-6
    F(2 on 3) = -2.995E-6

    This is a magnitude of 8.37E-6 N, which was incorrect
     
  7. Sep 4, 2013 #6

    gneill

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    The first force is wrong (direction).
    The second force is okay.
     
  8. Sep 4, 2013 #7
    The magnitude appears to be correct though, right? For some reason it's not
     
  9. Sep 4, 2013 #8

    rude man

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    Probably because I mixed you up. I suggested F(2 on 3 ) should be +, but it's -. Similarly, F(1 on 3) should be +, not -. But the magnitude of the difference is the same.

    Do it gneill's way which is right.
     
  10. Sep 4, 2013 #9
    In that case, F(1 on 3) is = 5.387E-6
    F(2 on 3) is = -2.995E-6

    Magnitude still appears to be 8.37E-6 N like you said. Am I using the wrong equation or something? This doesn't make sense to me
     
  11. Sep 4, 2013 #10

    gneill

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    Add the forces, then take the magnitude.
     
  12. Sep 4, 2013 #11
    Ohhh I gotcha, so 5.387E-6 + (-2.995E-6), which = 2.392E-6. So the magnitude is 2.392E-6N?
     
  13. Sep 4, 2013 #12

    gneill

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    Yup. Round to 3 significant figures for your final result.
     
  14. Sep 4, 2013 #13
    Perfect. Thank you all so, so much
     
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