# F10HW07 - 1 Problem 4.1 1.1 charge r q a = 2 q a = 2 = 2 q...

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Unformatted text preview: 1 Problem 4.1 1.1 charge r ' + q a = 2 + q a = 2 = 2 q a = 2 q a = 2 3 = 2 q l;m = qa l ¢ Y £ l;m ( = = 2 ; ' = 0) + Y £ l;m ( = = 2 ; ' = = 2) Y £ l;m ( = = 2 ; ' = ) Y £ l;m ( = = 2 ; ' = 3 = 2) £ = qa l ( 1) m s (2 l + 1)( l + m )! 4 ( l m )! 2 4 e i ( m )(0) | {z } 1 + e i ( m )( = 2) | {z } ( i ) m e i ( m )( ) | {z } ( 1) m e i ( m )(3 = 2) | {z } i m 3 5 P m l : cos( = 2) = qa l s (2 l + 1)( l + m )! 4 ( l m )! [1 + ( i ) m ( 1) m i m ] P m l (0) = qa l s (2 l + 1)( l + m )! 4 ( l m )! P m l (0) ( m even 2 2 i ( 1) ( m 1) = 2 m odd l = 1: q 1 ; 1 = q £ 1 ; 1 = qa r 3 2 (1 i ) l = 3: q 3 ; 1 = q £ 3 ; 1 = qa 3 r 7 (2 2 i ) 1 12 3 2 = qa 3 r 7 (1 i ) 1 4 q 3 ; 3 = q £ 3 ; 3 = qa 3 r 1260 (2 + 2 i ) 1 720 ( 15) = qa 3 r 35 (1 + i ) 1 4 1.2 charge r ' + q a 2 q q a 1 q l;m = q 2 4 a l Y £ l;m ( = 0 ; ' = 0) 2 : l Y £ l;m ( = 0 ; ' = 0) + a l Y £ l;m ( = ; ' = 0) 3 5 = qa l s (2 l + 1)( l m )! 4 ( l + m )! " e im (0) | {z } 1 P m l (cos(0)) + e im (0) | {z } 1 P m l (cos( )) # = qa l s (2 l + 1)( l m )! 4 ( l + m )! [ P m l (1) + P m l ( 1)] Because we have azumuthal symmetry, only the m = 0 terms survive. Noting that that P l (1) = 1, P l ( 1) = ( 1) l : q l;m = qa l r (2 l + 1) 4 ¢ 1 + ( 1) l £ = qa l r (2 l + 1) 4 ( 2...
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F10HW07 - 1 Problem 4.1 1.1 charge r q a = 2 q a = 2 = 2 q...

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