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Laurent Series Expansion of Electrostatic Potential
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[QUOTE="Mattkwish, post: 4520106, member: 489515"] [h2]Homework Statement [/h2] 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 first 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. [h2]Homework Equations[/h2] Ue = (2/4piε) * [(Q / √(x^2 + y^2)) - (Q / √(x^2 + y^2 + D^2))] [h2]The Attempt at a Solution[/h2] For part (b) specifically, i am completely lost as to how to arrange the equation when r >> D. [/QUOTE]
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Laurent Series Expansion of Electrostatic Potential
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