Can anyone tell me how to derive the equation for potential of an electric linear quadrupole?(adsbygoogle = window.adsbygoogle || []).push({});

Diagram of Quadrupole

My attempts yielded:

[tex]V = \frac{q}{4\pi\epsilon_0} [\frac{1}{R_1} - \frac{2}{R} + \frac{1}{R_2}][/tex]

[tex] V = \frac{q}{4\pi\epsilon_0} [\frac{RR_2 - 2R_1R_2 + RR_1}{RR_1R_2}] [/tex]

Approximating with R >> d, I set [tex]RR_1R_2 = R^3[/tex]

And I set [tex]R_1 = R-d\cos\theta[/tex] and [tex]R_2 = R+d\cos\theta[/tex]

Substituting in the equation, I get:

[tex] V = \frac{qd^2}{4\pi\epsilon_0R^3} [1+\cos{2\theta}] [/tex]

However, the book's answer states that the potential

[tex] V = \frac{qd^2}{8\pi\epsilon_0R^3} [1+3\cos{2\theta}] [/tex]

And I don't see how they got that. Was my approximation too inaccurate? Is there a geometric way to approximate this without going into taylor/maclaurin series expansion?

Thanks for any help offered!

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# Homework Help: Potential of a Linear Quadrupole

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