Laurent Series Expansion of Electrostatic Potential

  • Thread starter Mattkwish
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Homework Statement



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.

Homework Equations



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



The Attempt at a Solution



For part (b) specifically, i am completely lost as to how to arrange the equation when r >> D.
 

Answers and Replies

  • #2
TSny
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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.
 
  • #3
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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!
 
  • #4
TSny
Homework Helper
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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)
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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