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Potential of a Linear Quadrupole

  1. Feb 2, 2005 #1
    Can anyone tell me how to derive the equation for potential of an electric linear quadrupole?

    Diagram of Quadrupole

    My attempts yielded:

    [tex]V = \frac{q}{4\pi\epsilon_0} [\frac{1}{R_1} - \frac{2}{R} + \frac{1}{R_2}][/tex]

    [tex] V = \frac{q}{4\pi\epsilon_0} [\frac{RR_2 - 2R_1R_2 + RR_1}{RR_1R_2}] [/tex]

    Approximating with R >> d, I set [tex]RR_1R_2 = R^3[/tex]
    And I set [tex]R_1 = R-d\cos\theta[/tex] and [tex]R_2 = R+d\cos\theta[/tex]

    Substituting in the equation, I get:

    [tex] V = \frac{qd^2}{4\pi\epsilon_0R^3} [1+\cos{2\theta}] [/tex]

    However, the book's answer states that the potential

    [tex] V = \frac{qd^2}{8\pi\epsilon_0R^3} [1+3\cos{2\theta}] [/tex]

    And I don't see how they got that. Was my approximation too inaccurate? Is there a geometric way to approximate this without going into taylor/maclaurin series expansion?

    Thanks for any help offered!
     
  2. jcsd
  3. Feb 2, 2005 #2
    your reasoning seems fine and your algebra is correct..( except you have a typo, it should be 8pi in the denomenator instead of 4 pi)
    is that the exact problem? do you draw your graph or is it provided in the problem? I wanna make it clear before I get my hand wet......

    EDIT
    you said you have a linear quadrupole but I see 3 charge only.... that is not a quadrupole, that's why I asked this question.....
     
  4. Feb 2, 2005 #3
    Hi!

    Perhaps I should have made it clearer. It's a linear quadrupole because the charge in the middle (at the origin is -2q) while the charges on the top and the bottom are +q. The diagram itself was taken directly from the textbook, so it should be correct.

    Since it was badly drawn, though, I'll try to explain it a bit better. At the origin, there is a -2q charge. At distance +/-d from the origin, on either side of the -2q charge on the same axis, are two +q charges.

    As for the [tex]8\pi[/tex], can you tell me how you got that?

    In the second-to-last step in my algebra, I have:

    [tex]V = \frac{qd^2}{4\pi\epsilon_0R^3} [2\cos^2\theta] [/tex]

    Which then converted to [tex][1+\cos{2\theta}][/tex].

    Thanks a lot!
     
  5. Feb 2, 2005 #4

    dextercioby

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    In my mind,everything you did is perfect...Are u sure with the values of the charges.You know,when added,they give 0...

    Daniel.
     
    Last edited: Feb 2, 2005
  6. Feb 2, 2005 #5
    Are they not supposed to add up to 0? For example, in a dipole, you have +q and then -q, which adds up to zero.

    The textbook has been known to have errors, but I'm not sure my answer should be that far off from the given one. Does anyone have any links to a site that could give the equation for the potential of a linear quadrupole? I googled for it and only got an equation (scroll to #6) that was sort of close to mine which calculated for any point along the axis (instead of any point that could also be away from the axis).
     
  7. Feb 2, 2005 #6
    I've spend some time on this problem but can't find any error.....

    ps
    the 4 pie things, I thought it was a typo becasue i saw the answer is 8pie but indeed it is not....
     
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