Laurent Series Expansion of Electrostatic Potential

[Mattkwish]

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.

 

[TSny]

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.

 

[Mattkwish]

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!

 

[TSny]

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?

 

[paranormal]

i understand that in this limit, the term D^2 in the denominator can be neglected, but i am unsure how to proceed with the Laurent series expansion. Any help or guidance would be greatly appreciated. Thank you.


I would first like to commend you for your efforts in solving part (a) of the question and for seeking clarification on part (b). It shows your dedication and determination to understand the problem fully.

To address your question, in the limit r >> D, the term D^2 in the denominator can indeed be neglected. This simplifies the electrostatic potential to:

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

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

= 0

Therefore, in this limit, the electrostatic potential becomes zero. This makes sense intuitively because as r >> D, the charges are very far apart and their influence on the point P in the x, y-plane becomes negligible.

As for the Laurent series expansion, since the potential is zero in this limit, the first non-zero term would be the second term in the series. This would make the expansion:

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

= -(Q/2*pi*epsilon) * [1/√(x^2 + y^2)] + higher order terms

I hope this helps clarify your doubts and allows you to proceed with the Laurent series expansion. As for part (c) of the question, a linear quadrupole is called so because it consists of two positive charges and one negative charge arranged in a line, resembling the shape of a quadrupole. This arrangement of charges creates a dipole moment, which is a measure of the strength and direction of the electric field. In a linear quadrupole, the dipole moment is zero, while in a true quadrupole (four charges arranged in a square), the dipole moment is non-zero.

I hope this response has been helpful. Keep up the good work in your scientific pursuits!