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.
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.