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Electron vector analysis question

  1. Jun 17, 2008 #1
    3 electrons q1, q2, q3 setup in a right angled triangle

    q1
    |\
    | \
    q3-q2

    charges are

    q1 = +2.5 X 10^-17C
    q2 = +3.0 X 10^-17C
    q3 = +3.5 X 10^-17C

    distances between the charges
    q1-q3 = .03 m
    q1-q2 = .05 m

    setup FBD

    \
    \ 2F1
    \
    /
    / 3F1
    q1

    2F1 = k*q1*q2 / .05^2
    = 2.7 X 10^-21N
    3F1 = k*q3*q1 / .03^2
    = 8.75 X 10^-21N

    now I think I use cosine law to figure this out

    Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos135)
    = 7.1 X 10^-21

    with all charges being positive, all qs are experiencing repulsion so the direction is away from q1

    the question "calculate the magnitude and direction of force on q1"
     
  2. jcsd
  3. Jun 17, 2008 #2

    rl.bhat

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    To calculate the magnitude and direction of force on q1, find the force due to q3 on q1 and force due to q2 on q1. Find the resultant of these forces. Find the cosine of angle between these forces. It is not cos135. The direction of the resultant is given by angle made by the resultant with one of the forces.
     
  4. Jun 18, 2008 #3
    well, I believe I have found the forces acting on q1 (by q2 and q3). I guess my problem is I dont know what to do next. From the example I am given in the text they used cosine rule and a FBD to calculate the resultant of these forces. Im guessing my problem is my Trig is rusty......

    how do I determine what theta should be??? (in the book they use an isosceles and use 120degrees in this example its a right angle triangle).
     
  5. Jun 18, 2008 #4

    Kurdt

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    You have a right angled triangle and you know the length of two sides. That is all the information you need to calculate anything about that triangle. Can you remember the trigonometric ratios for a right angled triangle?
     
  6. Jun 18, 2008 #5
    sin0 = opp/hyp

    sin^-1(.03/.05) = 36.8 = 37 degrees
     
  7. Jun 18, 2008 #6

    Kurdt

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    You don't have the opposite, you have the adjacent.
     
  8. Jun 18, 2008 #7
    Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos37)

    ??

    I really need to touch up on my trig before I go back to school
     
  9. Jun 18, 2008 #8

    Kurdt

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    The net force will be the vector sum or resultant of the two vectors.
     
  10. Jun 18, 2008 #9
    sorry Im lost here, unfortunately the book provided for this distance ed course is poorly written and the examples provided dont really help with answering of questions.

    what in the trig am I messing up??

    seeing as this is based on the original distribution of forces, Im asuming Im working with the right angled triangle.

    Im trying to figure out theta for the upper left portion of this triangle (it probably has a name but I havnt touched trig in 10+ years). when looking for theta in this region, the triangle is setup as

    | = A
    \ = H
    __ = O

    q1 - q3 = A = .03m
    q1 - q2 = H = .05m

    given this information and the little trig I know, I can use cosine rule to solve for theta

    where cos = A/H

    cos^-1 (.03/.05) = 53 Degrees

    am I then using
    Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos53)

    Its absolutely absurd that I have nobody through the distance education program to ask questions and I am left to throw my self at the mercy of the internet forums.

    Thanks for any guidance
     
  11. Jun 18, 2008 #10

    Kurdt

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    Its just the final part of adding the forces now that you're not doing correctly. You're gonna have to set up a coordinate system and find the components of each vector along each of the axes. The following page may help you with that.

    http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
     
  12. Jun 18, 2008 #11
    so is this where I draw the FBD??

    and I guess I will need to know the angles of the other vectors

    so
    q1 - q2 = 53
    q3 - q1 = 90
    q2 - q3 = 180 - (53 + 90) = 37

    and

    2F1 = k*q1*q2 / .05^2
    = 2.7 X 10^-21N
    3F1 = k*q3*q1 / .03^2
    = 8.75 X 10^-21N
    2F3 = k*q2*q3 / .04^2
    =5.9 X 10^-21 N

    So with all this information I should be able to answer the question?????
     
  13. Jun 18, 2008 #12
    the reason I keep going back to this
    Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos0)

    is because thats how they answered the question in the text (solving trignometrically using cosine rule with the FBD)
     
  14. Jun 18, 2008 #13

    Kurdt

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    I get what you're doing now sorry. Its not often one uses the cosine rule to find resultant vectors. So you have the angle between the vectors with both starting from the same point. You need to find the angle between the vectors if you draw a diagram following the triangle law for addition of vectors. I take it you're familiar with that law if you're using the cosine rule to get the magnitude of resultant vectors. The other angle you worked out will help you.
     
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