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Homework Help: Electrostatic Force question

  1. Sep 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Three charges are fixed to an x, y coordinate system. A charge of +18 µC is on the y axis at y = +3.0 m. A charge of -16 µC is at the origin. Lastly, a charge of +45 µC is on the x axis at x = +3.0 m. Determine the magnitude and direction of the net electrostatic force on the charge at x = +3.0 m. Specify the direction relative to the -x axis.

    2. Relevant equations

    F = k*q1*q2 / r^2

    3. The attempt at a solution

    9e9 * 18e-6 * 16e-6 / 3^2 = .28768 N


    9e9 * 45e-6 * 16e-6 / 3^2 = .7192N

    (.29^2 + .72^2)^.5 = .7746N

    tan^-1 ( .29/.72) = 21.8 degrees...90 - 22 = 68 degrees below the x-axis

    Can anyone figure out where I am going wrong? I don't get what i am not seeing.

    Thank you!
  2. jcsd
  3. Sep 3, 2007 #2


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    This is wrong for calculating the net force... first find the net x-component of the force acting on the charge (so add the two components due to each charge)... then find the net y-component (only one charge needs to be considered here since the charge on the x-axis only contributes an x-component)...

    Then use the pythogrean theorem...
  4. Sep 3, 2007 #3
    If they are straight on the x-axis just out 3.0m and the other one straight up on the y-axis just up 3.0 m...then doesn't that make them 90 degree angles? meaning no x component on the charge on the y axis and then for the charge on the x-axis there is no y component?

    So, just a charge 3.0 m out at 0 degrees and one 3.0m up at 90 degrees?
    Thus only the .29 and .72 (clearly it is wrong, but i am not getting how to calculate otherwise).

  5. Sep 3, 2007 #4


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    The charge at (0,3) exerts both an x and y component on the charge at (3,0)...

    The question is asking for the force on the charge at (3,0)... not the force on the charge at (0,0)
  6. Sep 3, 2007 #5
    Ok so i was just not htinking in vectors...but here is what i have (i only have one more chance at this before it is due so i am trying to be cautious).

    K * 18e-6 * 16e-6 / 9 = .288 = y comp1
    K* 18e-6*45e-6/9 = .81 = x comp1

    K*45e-6*18e-6/9 = .81 = y comp 2
    K*45e-6*16e-6/9 = .72 = x comp2

    x1 + x2 = 1.53N
    y1+y2 = 1.098 N

    (1.098^2 + 1.53^2)^.5 = 1.88N

    THEN for the angle i am doing tan^-1 (1.098/1.53) = 35.66 degrees and relative to the -x axis that is -35.66 degrees if you draw the line to extend back through the origin and into quadrant 3?

    And thanks for all of your help so far :)
    It is definitely helping me past the road blocks!
  7. Sep 3, 2007 #6


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    You're making a common mistake calculating the components... Please look at post #22 in this thread:


    Also, I'm confused as to why you have a calculation for a force between the 18muC and -16muC charges... you only need the force on the 45muC charge...

    The force that the -16muC charge exerts on the 45muC charge is 0.72N in the -x direction... or -0.72 in the x direction... that part is right.

    Now you need to calculate the x and y components of the charge by the 18muC on the 45muC after looking at the post in the thread I linked...

    First calculate the net force of the 18muC on the 45muC. Then calculate the x component and y component.
  8. Sep 3, 2007 #7
    ok...it asks for: kq / x^2+y^2 * cos @ (where @ = theta).

    only thing i can think of is 45 degrees...so:

    kq / x^2 + y^2 * cos (45) = ((9e9 * 16e-6) / (3^2))* cos (45) + ((9e9 * 18e-6) / (3^2))* cos (45)?

    (currently working on this...i realize i forgot the x and y component part of the problem)
    Last edited: Sep 3, 2007
  9. Sep 3, 2007 #8


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    No, the x^2 + y^2 is in the denominator...

    What is the net force of the 18muC on the 45muC... what is r? what is r^2?
  10. Sep 3, 2007 #9
    bleh. sorry the problems are getting jumbled in my head...

    F = 9e9 * 45e-6 * 18e-6 / 9 = .81 N
    and we already determined .72 N

    So, now i calculate the x and y component?

    i get: x1 = 0, y1 = .81
    x2 = .72, y2 = 0

    pyth. theo. => (.81^2+.72^2)^.5 = 1.08N?
  11. Sep 3, 2007 #10


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    Why do you have 9 in the denominator?
  12. Sep 3, 2007 #11
    God i don't get this problem...

    i got all the starred, double starred, etc. problems...and this no star basic arse problem i just am not getting. Why on earth is this one so hard for me LOL

    Last edited: Sep 3, 2007
  13. Sep 3, 2007 #12
    9 in the denominator is from 3^2 as r^2

    the only other thing i have come up with is since it is the repulsive force...then 3 sqrt(2) so 18 in the denominator giving me .405

    .405 and .72 are my two sides then?
  14. Sep 3, 2007 #13


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    Yes, it should be 18 in the denominator... I'm not understanding why you say "since it is the repulsive force"... whether it was attractive or repulsive... the distance is 3sqrt(2) hence r^2 = 18... Hope I don't seem like a jerk... just want to make sure you understand this part. :smile:

    No wait... 0.405 N is the net force exerted by the charge at (0,3) on the charge at (3,0)... that's right... now what is the x-component of this force... and what's the y-component of this force...
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