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Coulombs Law with one unknown charge

  1. Jan 25, 2005 #1
    y(m)
    /\
    2 | q3
    |
    |
    1 q1
    |
    |
    ------q2--> x(m)
    2
    *graphic* is kind of distorted.
    So the coordinates of the charges are:

    q1 = (0,1)
    q2 = (2,0)
    q3 = (2,2)

    given:
    q1 = 2.5 x 10^-5 C
    q2 = 2.0 x 10^-5 C
    q3 = ?

    Problem:
    If the force on q1 points in the -x direction
    (a) what is q3.
    (b) what is the magnitude of the force on q1.

    Notation Key:
    Why? Sorry, but the LaTeX seems to be turned off or something.
    So I hope this is ok.

    r_ab = distance from a to b
    R_ab = unit vector from a to b

    Coulombs law:
    F_ab=k q_a q_b
    --------- R
    C

    r_13 = sqrt(1+2^2) = sqrt(5)m
    r_23 = 2m

    R_23 = -J = <0,-1>
    R_13 = (-2I-J)/sqrt(5) = <(-2sqrt(5))/5,-sqrt(5)/5>

    F_net_3 = F_23 + F_13 from superposition principle

    F_23 = (9 x 10^9 N m^2) (2.0 x 10^-5 C) q3 C R_23
    ---------------- ---------------
    C^2 (2m)^2

    F_13 = (9 x 10^9 N m^2) (2.5 x 10^-5 C) q3 C R_23
    ---------------- ---------------
    C^2 (2m)^2

    F_23 = <0, -1 * (1.8 x 10^5 q3)/5> N
    F_13 = <(2.25 x 10^5 q3)/5 (-2)sqrt(5)/5
    ,2.25 x 10^5 q3)/5 (sqrt(5))/5>

    F_23 = <0, -3.6 x 10^4 q3> N
    F_13 = <-40249.2 q3, -20124.6 q3> N


    Now this is where I get stuck since I have to many unknowns.
    I just don't know what to do from here... ANY help would be
    amazing. Thanks.
     
    Last edited: Jan 25, 2005
  2. jcsd
  3. Jan 25, 2005 #2
    your r_13 and r_23 is completely wrong already, I didn't check the rest....
    do you having difficulty finding distance?
     
  4. Jan 25, 2005 #3
    Sorry about that. Actually the mistake was in the coordinate system. q3 is supposed to be (2,2) not (0,2) like I had... so I guess I have trouble typing in proper numbers, not finding the distance :)

    P.S. It looks god awefull without latex. Is it turned off or something?
     
    Last edited: Jan 25, 2005
  5. Jan 25, 2005 #4
    draw the graph, use symmetry argue q3 is equal to q2(hopefully you can see that), it will save you a lot of time
    then find the x component of the force (should be easy for you)
     
  6. Jan 25, 2005 #5
    I don't really see why q3 is equal to q2. But, I'll just go under the assumption that it is and work though the process.
    Thank you for the help :)
     
  7. Jan 25, 2005 #6

    learningphysics

    User Avatar
    Homework Helper

    The y-components of the two forces have to cancel, since the force at q1 is in the -x direction.

    Write an equation for the sum of the 2 y-components and set it equal to zero. Then you'll see that q3=q2.
     
  8. Jan 26, 2005 #7
    Thank you guys so much. I guess I was just making the problem harder then it was. I definitely see why the forces cancel if q3=q2. Ok cool... :) Just glad I got that done.
    By the way, this board is awesome.
     
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