15_2_15 - 15. The two-dimensional dipole of strength has...

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15. The two-dimensional dipole of strength μ has potential φ( x , y ) = lim 0 m = μ m 2 " ln x 2 + ± y - 2 ² 2 ! - ln x 2 + ± y + 2 ² 2 !# = μ 2 lim 0 ln x 2 + ± y - 2 ² 2 ! - ln x 2 + ± y + 2 ² 2 ! (apply l’Hˆopital’s Rule) = μ 2 lim 0 - ± y - 2 ² x 2 + ± y - 2 ² 2 - ± y + 2 ² x 2 + ± y + 2 ² 2 = - μ y x 2 + y 2 = - μ y r 2 . Now ∂φ x =
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