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Homework Help: Calculate the direction and magnitude problem

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    3 charges are arranged in the diagram below. their magnitudes are:
    q1=2.5*10^-17C
    q2=3.0*10^-17C
    q3= 3.5*10^-17

    calculate the direction and magnitude of q1
    2. Relevant equations
    F=kq^2/r^2


    3. The attempt at a solution
    well, i used f=kq^2/r^2, the charge of q3= (9*10^9)(3.5*10^-17)^2/(0.5)^2= 1.2*10^-17
    q2=(9*10^9)(3*10^-17)^2/(0.50)^2= 3.24*10^-17N
    im unable to find the angle of q1, how would i solve it really confused, i know you can use cos law but angle unknown, thank you
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2010 #2
    Re: electrostatic

    btw , the angle from q3 to q1 is 90 deg
     
  4. Feb 21, 2010 #3

    Redbelly98

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    Re: electrostatic

    I'll assume you mean the direction and magnitude of the force on q1 :smile:
    That equation is wrong. Instead, use
    F = k qa qb / r2
    for the force between two charges qa and qb
     
  5. Feb 21, 2010 #4
    Re: electrostatic

    sorry, i just minimized the formula since it was late, i know im suppose to use
    F=q1q2/r^2 ( but how do i do this, im really confused) this question , could it be resolved in components ?
     
  6. Feb 21, 2010 #5

    Redbelly98

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    Re: electrostatic

    After you calculate F=kq1q2/r^2, and think about what direction that force is acting, then yes you resolve it into components.
     
  7. Feb 21, 2010 #6
    Re: electrostatic

    so do i calculate the charge from q1-q3, and q1-q2 :S, sorry the question is really bothering me
     
  8. Feb 21, 2010 #7

    Redbelly98

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    Re: electrostatic

    You calculate the force from q3 on q1, and the force from q2 on q1.

    You don't have to calculate any charge, the charge values are provided in the problem statement. The problem is to calculate force, not charge. They are two different things. Force is not charge. Force is not charge. Force is not charge. Force is not charge.
     
  9. Feb 21, 2010 #8
    Re: electrostatic

    ok i think i got it, but how would you draw a diagram if the one charge was attractive, in this case all the charges are repulsive
     
  10. Feb 21, 2010 #9
    Re: electrostatic

    You're calculating the force not the charge. The force q1-q3 is:

    [tex] F = k_e \frac{q_1 q_3}{r^2} [/tex]

    The force q1-q2 similarly.

    [tex] F = k_e \frac{q_1 q_2}{r^2} [/tex]

    You can find the angle in q1 corner using the tangent function. Now sum them like vectors and you're done.
     
  11. Feb 21, 2010 #10
    Re: electrostatic

    thank you very much for the help, i'm able to get it
     
  12. Feb 21, 2010 #11
    Re: electrostatic

    okay , so using the pythagorean theorem, since hypotenuse is known it would be
    b^2=a^2-c^2 ( then im done) :) , if that's all i thank you all for the help
     
  13. Feb 21, 2010 #12

    Redbelly98

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    Re: electrostatic

    I don't really understand that last post, but if you show your work we can help tell you if it's right or wrong.
     
  14. Feb 21, 2010 #13
    Re: electrostatic

    okay, the charge of 1 to charge 3 is (9*10^9)(3.5*10^-17)(2.5*10^-17)/(0.3)^2
    = 8.75*10^-23N
    then, charge 1 to 2 is (9*10^9)(3*10^-17)(2.5*10^-17)/(0.5)^2 = 2.7*10^-23
    after i have these 2 charges i have to do pythogorean theroem but for the unknown side im getting a negative root ans
     
  15. Feb 21, 2010 #14

    Redbelly98

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    Re: electrostatic

    Okay, we are starting to get somewhere now. But something is rather confusing. In the drawings you show in Post #1, the distance between charges 1 and 3 is not 0.3m. The distance between charges 1 and 2 is not 0.5m. But these are the distances you used in the calculations of the forces.

    Question: which of the two figures you showed earlier is the correct configuration of the charges?

    Finally, if I may politely ask you ... please stop using the word "charge" when you really mean (or should mean) "force". It indicates that you are not really paying attention to what you are doing or what we are saying.
     
  16. Feb 21, 2010 #15
    Re: electrostatic

    ok , sorry about that, but im getting it a little more the correct configuration is the one to the left. the reason i drew the one to the right is to show the repulsive forces that q1 experiences.
     
  17. Feb 21, 2010 #16

    Redbelly98

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    Re: electrostatic

    Okay, thanks for clearing that up. You'll need to recalculate the forces using the correct distances between the charges.
     
  18. Feb 21, 2010 #17
    Re: electrostatic

    no problem, i thought i already calculated the force b/w q1 and q2, likewise i calculated the force between q1 & q3 ,would'nt i need to draw a diagram showing the repulsive forces on q1( the picture to the right) ?
     
  19. Feb 22, 2010 #18

    Redbelly98

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    Re: electrostatic

    Reread my post #14, especially the part about the distances between the charges.
    I thought the picture to the right was incorrect? The pictures can't both be correct, since they show the charges in different arrangements.

    I have to say, you do need to be careful and pay more attention, otherwise there is no way I can help you out.
     
  20. Feb 22, 2010 #19
    Re: electrostatic

    OH MAN, thank i completely messed up the distances
     
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