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Laurent Series Expansion of Electrostatic Potential

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a series of three charges arranged in a line along the z-axis, charges +Q at
    z = D and charge -2Q at z = 0.
    (a) Find the electrostatic potential at a point P in the x, y-plane at a distance r from
    the center of the quadrupole.
    (b) Assume r >> D. Find the first two non-zero terms of a Laurent series expansion
    to the electrostatic potential you found in the fi rst part of this problem.
    (c) A series of charges arranged in this way is called a linear quadrupole. Why?

    I have already solved part (a) of the question, and found the electrostatic potential, Ue, to be

    Ue = (2/4*pi*epsilon) * [(Q / \sqrt(x^2 + y^2)) - (Q / \sqrt(x^2 + y^2 + D^2))]

    My next step was to divide by D^2 out of the square root so i can put the equation in a form that fits to a pre-solved integral sheet that we were handed.

    2. Relevant equations

    Ue = (2/4piε) * [(Q / √(x^2 + y^2)) - (Q / √(x^2 + y^2 + D^2))]



    3. The attempt at a solution

    For part (b) specifically, i am completely lost as to how to arrange the equation when r >> D.
     
  2. jcsd
  3. Sep 29, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    Hello Mattkwish and welcome to PF!

    Try rewriting your expression for the potential in terms of r and D instead of x, y, and D.
    Then think about how to make an approximation for r >> D.
     
  4. Sep 30, 2013 #3
    Hi,

    I attempted to substitute in r for x and y and established the substitution of (r0)^2 = x^2 + y^2. I then substituted in for r0, and was able to see that the bottom of one of the terms tends towards 0. So now i have the equation:

    Ue = 2Q / 4piε * [1 - 1/r0] (the 1 / √r^2 + D^2 term goes towards 1/1)

    However now i am lost as to how to construct a Laurent Series from this equation. My next step would be to use the equation (1/n!) * (d^nF / dz^n) |z=a and a similar one for the coefficients, however i am struggling with this. I remember using an integral to find the power series, which i could do for this function... could use some advice!
     
  5. Sep 30, 2013 #4

    TSny

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    Gold Member

    This doesn't look right. Note that you have [1 - 1/r0]. The 1 has no dimensions while the 1/r0 has dimensions of 1/length. So, there is an inconsistency that shows something is wrong.

    Let's go back to your expression

    ##V_e = (2/4\pi\epsilon) (Q / \sqrt{x^2 + y^2} - Q / \sqrt{x^2 + y^2 + D^2})##

    I took the liberty of using ##V## for potential instead of ##U##, since that is the more common notation. Of course you can simplify the 2/4 and also factor out Q. What does the expression become after writing it in terms of ##r## but before making any simplifications?
     
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