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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

Ask a homework question - tutors are online