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Homework Help: Electric Field in a point

  1. Sep 4, 2007 #1
    I want to find an electric field in a point (x,y,z) generated by 2 charges:
    q1 = q, and q2=2q, q1’s position = (1,2,3) and q2’s position = (4,5,6),

    I tried to solve it by using superposition principle, adding the E of each charge to point x,y,z, I am doing something wrong probably with vectors calculation.

    Please explain.
     
  2. jcsd
  3. Sep 4, 2007 #2

    G01

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    Where in the calculation are you getting stuck? I'm not going to be able to give you any decent help if you don't show some work.
     
  4. Sep 4, 2007 #3
    well,
    E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1 + q/r2^2 * r2)
    r1 and r2 vectors
     
  5. Sep 4, 2007 #4
    r1=(x-1,y-2,z-3)
    r2=(x-4,y-5,z-6)

    after that and calculating all constants i get stuck.
     
  6. Sep 4, 2007 #5

    learningphysics

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    The vectors should be unit vectors...
     
  7. Sep 4, 2007 #6
    yes unit vector,
    I get: r1/|r1| but this doesn’t give me anything
     
  8. Sep 4, 2007 #7
    If it should be like this:

    E[x] = E1*(1/sqrt(14),0,0)+E2..

    ?
    Thanks.
     
  9. Sep 4, 2007 #8

    learningphysics

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    Yeah, how is the answer supposed to look? How do you know your answer is wrong?
     
  10. Sep 4, 2007 #9
    I don't know the answer, but my answer has form with a lot of x,y and z variables.
     
  11. Sep 4, 2007 #10

    learningphysics

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    There should be 2q in the numerator or r2^2.
     
  12. Sep 4, 2007 #11
    well,
    E = 1/[4*Pi*epsilon0] * (q/r1^2 * r1/|r1| + q/r2^2 * r2/|r2|)
    r1 and r2 vectors

    did you solve this one?
    You would get lots of x,y,z vars, if you do

    yea, changing to q2=2q should give the answer..
     
  13. Sep 4, 2007 #12
    Yes, 2q.

    I have a follow up question, to find field in (10,10,10) where q1=10^-6 q2=2*10^-6
    and the answer for that is 10^-9K(21.09x’+17.76y’+14.43z’)

    My answer and the point (10,10,10) doesn’t give this result.
     
  14. Sep 4, 2007 #13
    rootX,
    yes I get a mess of x,y and z.
     
  15. Sep 4, 2007 #14
    I am doing something wrong with vectors, it's shouldn't be complicated...
     
  16. Sep 4, 2007 #15

    learningphysics

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    Just leave it as the sum of the two fields... also you can factor out [tex]\frac{q}{({r_1^2})^{3/2}}[/tex] for the first part... and [tex]\frac{q}{({r_2^2})^{3/2}}[/tex] for the second part...

    so all the messy stuff in the denominator can get factored out of the vectors... don't actually multiply out (x-a)^2 + (y-b)^2.... etc.
     
    Last edited: Sep 4, 2007
  17. Sep 4, 2007 #16
    I get
    kq (0.123,1.17798..)
    I guess I also messed up something..
     
  18. Sep 4, 2007 #17
    ok, I am 55% sure that this answer is wrong: 10^-9K(21.09x’+17.76y’+14.43z’)

    here's what I got from maxima:
    kq(0.14227258171877 , 0.20662308551546 , 0.270973589312)
     
  19. Sep 4, 2007 #18

    learningphysics

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    I'm getting the answer that is given: 10^-9K(21.09x’+17.76y’+14.43z’)
     
    Last edited: Sep 4, 2007
  20. Sep 4, 2007 #19
    learningphysics,
    can you please give your solution?
     
  21. Sep 4, 2007 #20

    learningphysics

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    Basically I got:

    [tex]kq*[\frac{(9,8,7)}{{194}^{3/2}} + 2*\frac{(6,5,4)}{{77}^{3/2}}][/tex]

    this evaluates to the given answer.
     
    Last edited: Sep 4, 2007
  22. Sep 4, 2007 #21
    lol
    I was doing [sqrt(194)]^3/2 and [sqrt(77)]^3/2
    Thanks
     
  23. Sep 4, 2007 #22
    and I was using (1,2,3) for (9,8,4) and (4,5,6) for..
     
  24. Sep 5, 2007 #23
    Another related question:

    There are 2 linear charges distributed on y and x axis, segments lengths 0<=x<=l and 0<=y<=l.
    Density of the charges is not uniformed: gamma(x)=bx and gamma(y)=by.
    Need to fine E(0,0,z)

    I tried to do following:

    Because those 2 segments lay on axis x and axis y the needed Electric field is on axis z.
    And because of symmetric I can calculate the contribution of one segments and contribution of the second will be the same.

    So I need to find dE and then dE_z.

    dE= [tex]\frac{kdq}{r^2}[/tex]

    dE_z= [tex]\frac{kdq}{r^2}[/tex] sin ([tex]\alpha[/tex])

    r^2 = [tex]\sqrt{z^2+l^2}[/tex]

    Am I doing right so far?
     
  25. Sep 5, 2007 #24

    learningphysics

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    You meant:

    [tex]r=\sqrt{z^2+l^2}[/tex]

    Yes, what you've described will give you the field for one segment at the point z along the z-axis... then you can double it to get the field along the z-axis...

    But you also need another component of the field at (0,0,z)
     
  26. Sep 5, 2007 #25
    What do you mean by another component?
     
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