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Unformatted text preview: Physics 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel Problem Set 3 Problem 2.7 5 Points a) : Greens function: Using cartesian coordinates x = ( x,y,z ), it is G ( x , x ) = 1 p ( x x ) 2 + ( y y ) 2 + ( z z ) 2 1 p ( x x ) 2 + ( y y ) 2 + ( z + z ) 2 . (1) Note the required symmetry in x and x . A coordinatefree form is G ( x , x ) = 1  x x  1  x x + 2 z ( z x )  . (2) b) : It is ( x ) = 1 4 Z x = x = Z y = y = z G ( x , x )  z =0 V ( x ) dx dy . (3) Using z G ( x , x )  z =0 = 2 z ( x x ) 2 +( y y ) 2 + z 2 3 and rewriting into cylindrical coordinates, x = cos x = cos y = sin y = sin (4) yields z G ( x , x )  z =0 = 2 z p z 2 + 2 + 2 2 (cos cos + sin sin ) 3 = 2 z p z 2 + 2 + 2 2 cos(  ) 3 . (5) The potential thus is ( ,,z ) = V z 2 Z 2 =0 Z a =0 d d p z 2 + 2 + 2 2 cos(  ) 3 . (6) c) : On the zaxis it is = 0, and (0 , * ,z ) = V z Z a =0 d d p z 2 + 2 3 = V z " 1 p z 2 + 2 # a = V (1 z z 2 + a 2 ) . (7) d) : We write ( ,,z ) = V z 2 ( 2 + z 2 ) 3 / 2 Z 2 =0 Z a =0 1 + 2 2 cos(  ) 2 + z 2  3 / 2 d d . (8) and expand (1 + ) 3 / 2 = 1 {z} + 3 2  {z } + 15 8 2  {z } ... A B C , (9) where = 2 2 cos(  ) 2 + z 2 1. Integration term by term yields A = Z 2 =0 Z a =0 d d = a 2 (10) B = 3 2 Z 2 =0 Z a =0 2 2 cos(  ) 2 + z 2 d d = 3 a 4 4( 2 + z 2 ) , (11) where the costerm integrates to zero. C = 15 8 Z 2 =0 Z a =0 4 4 3 cos(  ) + 4 2 2 cos 2 (  ) ( 2 + z 2 ) 2 d d = 5 a 6 8( 2 + z 2 ) 2 + 15 2 a 4 8( 2 + z 2 ) 2 , (12) where we have used that R 2 =0 cos(  ) d = 0 and R 2 =0 cos 2 (  ) d = . Collecting the terms into Eq. 8, it is found ( ,z ) = V a 2 z 2( 2 + z 2 ) 3 / 2 1 3 a 2 4( 2 + z 2 ) + 5 a 4 8( 2 + z 2 ) 2 + 15 a 2 2 8( 2 + z 2 ) 2 + ... , q . e . d . (13) Onaxis, the expression reduces to ( = 0 ,z ) = V a 2 2 z 2 1 3 a 2 4 z 2 + 5 a 4 8 z 4 ... , (14) while the expansion of the result of part c) for large z is (0 ,z ) = V (1 z z 2 + a 2 ) = V 1 1 + a z 2  1 / 2 ! V a 2 2 z 2 1 3 a 2 4 z 2 + 5 a 4 8 z 4 . (15) The results in Eq. 14 and Eq. 15 agree, as expected. Problem 2.8 5 Points a) : It is to be shown that the equipotential surfaces of two parallel line charges of equal magnitude and opposite polarity are cylinders. Using the variables identified in the figure, we claim that for any potential V there exists a cylinder with radius r and axis location identified by x such that for all values of the potential on the cylinder is V . We prove the claim by finding a unique solution for....
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This test prep was uploaded on 02/18/2008 for the course PHYS 350 taught by Professor None during the Spring '08 term at San Diego State.
 Spring '08
 none
 Magnetism

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