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Dipole moment for a charge arrangement

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the dipole moment for the following charge arrangement?

    (Please view attachment)
    http://img378.imageshack.us/my.php?image=dipoleal6.jpg

    [​IMG]



    2. Relevant equations


    p = [tex]\int r' \rho (r) d \tau '[/tex]

    3. The attempt at a solution

    also p = [tex]\sum qi r'i}[/tex]
     

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    Last edited: Nov 17, 2008
  2. jcsd
  3. Nov 17, 2008 #2

    gabbagabbahey

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    Your attachment is pending approval....it would be quicker if you were to upload the image to a free hosting site like imageshack.us and then just post a link to it.
     
  4. Nov 17, 2008 #3
  5. Nov 17, 2008 #4

    gabbagabbahey

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    Well, you have a collection of point charges, so I would use your second formula:

    [tex]\vec{p}=\sum q_i \vec{r'_i}[/tex]

    Since the total charge is zero, you are free to choose any point in space is your origin (when the total charge of a configuration is non-zero, the dipole moment depends on the choice of the origin)...so I would choose a coordinate system that makes expressing your [itex]\vec{r'_i}[/itex]s as easy is possible....any ideas on what that might be?
     
  6. Nov 17, 2008 #5
    Will spherical be the easiest one?
     
  7. Nov 17, 2008 #6

    gabbagabbahey

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    I would use Cartesian....my question is really: where do you want to put your x and y axes?
     
  8. Nov 17, 2008 #7
    i am not sure about the choice of axes and and their significance. Suggestion will be appreciated.
     
  9. Nov 17, 2008 #8

    gabbagabbahey

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    Well, I would put the origin at the point of intersection of the line segments in your diagram....the reason for this is that the distance to each of the charges from this point is very easy to calculate....Similarly, I would orient your coordinate system so that the x-axis point from this origin to the charge on the right in your diagram, and the y-axis points from the origin to the upper charge in your diagram....the reason for this is so that the directions of the vectors from the origin to each charge are just your unit vectors [itex]\pm \hat{x}[/itex] and [itex]\hat{y}[/itex]...This makes it very easy for you to express each of your [itex]\vec{r'_i}[/itex] and hence very easy to find the vector sum [tex]\vec{p}=\sum q_i \vec{r'_i}[/tex].....do you follow?

    What do you get for your dipole moment when you do this?
     
  10. Nov 18, 2008 #9
    Thank you! But when you add up charges, it goes to zero. So, how does it affect the net dipole moment ?
     
  11. Nov 18, 2008 #10

    gabbagabbahey

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    The net charge is zero, but the dipole moment isn't....as I said before, the fact that the net charge is zero allows you to choose any point as your origin when computing the dipole moment.....if the net charge were non-zero, the dipole moment would depend on which choice you made for the origin.

    BTW are you also Fourier X, or do you have the same question?
     
  12. Nov 18, 2008 #11
    I was just going through the problem and started wondering. Thanks anyway!
     
  13. Nov 18, 2008 #12
    if l is the distance of charge 2q in the y-axis (following the axes we assumed),
    p = -qa + -(qa) + 2ql

    am i in the right track ?
     
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