Dipole moment for a charge arrangement

  • Thread starter FourierX
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  • #1
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Homework Statement



What is the dipole moment for the following charge arrangement?

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

http://img378.imageshack.us/my.php?image=dipoleal6.jpg



Homework Equations




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

The Attempt at a Solution



also p = [tex]\sum qi r'i}[/tex]
 

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Answers and Replies

  • #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
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?
 
  • #5
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Will spherical be the easiest one?
 
  • #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?
 
  • #7
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i am not sure about the choice of axes and and their significance. Suggestion will be appreciated.
 
  • #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?
 
  • #9
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Thank you! But when you add up charges, it goes to zero. So, how does it affect the net dipole moment ?
 
  • #10
gabbagabbahey
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Thank you! But when you add up charges, it goes to zero. So, how does it affect the net dipole moment ?
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?
 
  • #11
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I was just going through the problem and started wondering. Thanks anyway!
 
  • #12
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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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