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Physical Quadrapole

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Can anyone verify what a physical quadrapole looks like? I'm trying to do a multipole expansion on one


    2. Relevant equations
    I understand what a physical dipole is and how to do a multipole extension on one


    3. The attempt at a solution
    My best guess is that a physical quadrapole is as follows...

    -q---+q

    +q----q

    with equal distance, d, between each charge? Could someone verify that this is correct
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    gabbagabbahey

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    Looks fine to me!:approve:

    As you would expect, the total charge is zero, so it has no monopole moment, and you should easily be able to show that the total dipole moment is also zero. Therefor, the potential should be dominated by the quadrapole moment, making it a good model for a physical quadrapole.

    If you are studying from Griffiths, he actually gives a short discussion (and a figure!) in the last paragraph of section 3.4.2.
     
  4. Oct 11, 2009 #3
    I read it over and thought it was probly that but I don't think he ever labels it as a physical quarapole.

    Best method to achieve the full expansion would be to treat the problem as 2 "individual" dipoles with the same theta, and then use the law of cosines, etc...?
     
  5. Oct 11, 2009 #4

    gabbagabbahey

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    That's one idea; just be careful to treat the dipoles as physical dipoles, not ideal dipoles and take note of the fact that they can't both be centered at the origin.

    Alternatively, you might choose your origin to be at the center of the charge distribution, and then just write down the potential due to the 4 point charges, and then Taylor expand your expression for large [itex]r[/itex]....Are you asked to find an expression for every term in the expansion, or just up to a certain order in 1/r?
     
  6. Oct 11, 2009 #5
    To answer your question, I'm to "calculate the potential energy of a physical linear electric quadrapole"

    And in response, I wasn't thinking about the origin being at the center of the quadrapole versus the the center of the dipoles (if I were to do it that way). But if I follow, I can simply do a multipole expansion for each of the charges, right?
     
  7. Oct 11, 2009 #6

    gabbagabbahey

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    Why do any multipole expansion at all? You are asked to find the potential energy of a physical quadrapole; surely you know how to calculate the exact potential energy of 4 point charges in the given arrangement?
     
  8. Oct 12, 2009 #7
    Errrr!!! Yea, no qualification like [tex]r\geq\geq d[/tex].... so just find the potential of each point and add them...

    The equation for potential is...

    [tex]V=\left(\frac{q}{4\pi\epsilon_0\cdot r}\cdot\hat{r}\right)[/tex]

    I'm tired, I'll look at this more in the morning, something to have nightmares about tonight... yay!! :)
     
  9. Oct 12, 2009 #8

    gabbagabbahey

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    Get some sleep!:zzz:

    Potential is a scalar, not a vector. There is no [itex]\mathbf{\hat{r}}[/itex] in the equation.:wink:
     
  10. Oct 12, 2009 #9
    Duh... sry lol... so i get...

    [tex]
    V=\left(\frac{q}{4\pi\epsilon_0}\right)\cdot\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}+\frac{1}{r_{3}}-\frac{1}{r_{4}}\right)
    [/tex]

    Like to reduce this into terms of d somehow... working on this, I'll see what comes out, but geometry never was my strong point
     
  11. Oct 12, 2009 #10
    ooo and to solve this... set r as the distance from the center to the point, P. Using the law of cosines.... it appears as if I'll have the answer in 2 different angular terms, like [tex]\theta[/tex] and [tex]\phi[/tex]... let's see what happens
     
  12. Oct 12, 2009 #11
    And that's exactly what happens... so I've got the equations for [tex] r_{1} , r_{2} , r_{3} , r_{4}[/tex] in terms of [tex]\theta[/tex] and [tex]\phi[/tex], how should I get these into a single term of the angle or can I?
     
  13. Oct 12, 2009 #12

    D H

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    No. Look at it this way. With this particular setup, those four lobes result from those four alternating positive and negative charges. There is no getting rid of that azimuthal dependency. There's no getting rid of that zenith dependency, either.
     
  14. Oct 12, 2009 #13
    Yea didn't think so, thank you. How do I mark a question resolved?
     
  15. Oct 12, 2009 #14

    gabbagabbahey

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    Don't forget that the question asks for the potential energy of the configuration, not just the potential.:wink:
     
  16. Oct 12, 2009 #15
    urrrrrrggh so my entire line of thought was for nothing and the equation becomes

    [tex]U=\left(\frac{1}{4\pi\epsilon_0}\right)\left(\frac{q_{1}q_{2}}{d}+\frac{q_{1}q_{3}}{d}+\frac{q_{1}q_{4}}{\sqrt{2} d}+\frac{q_{2}q_{3}}{\sqrt{2} d}+\frac{q_{2}q_{4}}{d}+\frac{q_{3}q_{4}}{d}\right)[/tex]

    Final Answer: [tex]U=\left(\frac{q^2}{4\pi\epsilon_0\cdot d}\right)\left(\frac{2}{\sqrt{2}}-4\right)[/tex]
     
  17. Oct 12, 2009 #16
    I can do that right?
     
  18. Oct 14, 2009 #17

    gabbagabbahey

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    Looks good to me!:approve:
     
  19. Oct 14, 2009 #18
    OK thanks for the heads up :). NOW how do I mark a problem resolved?
     
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