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midterm2

# midterm2 - Physics 505 Electricity and Magnetism Prof G...

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Physics 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel Problem Set 3 Problem 2.7 5 Points a) : Green’s function: Using cartesian coordinates x = ( x, y, z ), it is G ( x , x 0 ) = 1 p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z - z 0 ) 2 - 1 p ( x - x 0 ) 2 + ( y - y 0 ) 2 + ( z + z 0 ) 2 . (1) Note the required symmetry in x and x 0 . A coordinate-free form is G ( x , x 0 ) = 1 | x - x 0 | - 1 | x - x 0 + 2ˆ z z · x 0 ) | . (2) b) : It is Φ( x ) = 1 4 π Z x 0 = x 0 = -∞ Z y 0 = y 0 = -∞ ∂z 0 G ( x , x 0 ) | z =0 V ( x 0 ) dx 0 dy 0 . (3) Using ∂z 0 G ( x , x 0 ) | z 0 =0 = 2 z ( x - x 0 ) 2 +( y - y 0 ) 2 + z 2 3 and re-writing into cylindrical coordinates, x = ρ cos φ x 0 = ρ 0 cos φ 0 y = ρ sin φ y 0 = ρ 0 sin φ 0 (4) yields ∂z 0 G ( x , x 0 ) | z =0 = 2 z p z 2 + ρ 2 + ρ 0 2 - 2 ρρ 0 (cos φ cos φ 0 + sin φ sin φ 0 ) 3 = 2 z p z 2 + ρ 2 + ρ 0 2 - 2 ρρ 0 cos( φ - φ 0 ) 3 . (5) The potential thus is Φ( ρ, φ, z ) = V z 2 π Z 2 π φ 0 =0 Z a ρ 0 =0 ρ 0 0 0 p z 2 + ρ 2 + ρ 0 2 - 2 ρρ 0 cos( φ - φ 0 ) 3 . (6) c) : On the z -axis it is ρ = 0, and

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Φ(0 , * , z ) = V z Z a ρ 0 =0 ρ 0 0 0 p z 2 + ρ 0 2 3 = - V z " 1 p z 2 + ρ 0 2 # a 0 = V (1 - z z 2 + a 2 ) . (7) d) : We write Φ( ρ, φ, z ) = V z 2 π ( ρ 2 + z 2 ) 3 / 2 Z 2 π φ 0 =0 Z a ρ 0 =0 1 + ρ 0 2 - 2 ρρ 0 cos( φ - φ 0 ) ρ 2 + z 2 - 3 / 2 ρ 0 0 0 . (8) and expand (1 + ² ) - 3 / 2 = 1 |{z} + - 3 2 ² | {z } + 15 8 ² 2 | {z } - ... A B C , (9) where ² = ρ 0 2 - 2 ρρ 0 cos( φ - φ 0 ) ρ 2 + z 2 ¿ 1. Integration term by term yields A = Z 2 π φ 0 =0 Z a ρ 0 =0 ρ 0 0 0 = πa 2 (10) B = - 3 2 Z 2 π φ 0 =0 Z a ρ 0 =0 ρ 0 2 - 2 ρρ 0 cos( φ - φ 0 ) ρ 2 + z 2 ρ 0 0 0 = - 3 πa 4 4( ρ 2 + z 2 ) , (11) where the cos-term integrates to zero. C = 15 8 Z 2 π φ 0 =0 Z a ρ 0 =0 ρ 0 4 - 4 ρρ 0 3 cos( φ - φ 0 ) + 4 ρ 2 ρ 0 2 cos 2 ( φ - φ 0 ) ( ρ 2 + z 2 ) 2 ρ 0 0 0 = 5 πa 6 8( ρ 2 + z 2 ) 2 + 15 πρ 2 a 4 8( ρ 2 + z 2 ) 2 , (12) where we have used that R 2 π φ 0 =0 cos( φ - φ 0 ) 0 = 0 and R 2 π φ 0 =0 cos 2 ( φ - φ 0 ) 0 = π . Collecting the terms into Eq. 8, it is found Φ( ρ, z ) = V a 2 z 2( ρ 2 + z 2 ) 3 / 2 1 - 3 a 2 4( ρ 2 + z 2 ) + 5 a 4 8( ρ 2 + z 2 ) 2 + 15 a 2 ρ 2 8( ρ 2 + z 2 ) 2 + ... , q . e . d . (13) On-axis, the expression reduces to Φ( ρ = 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 Φ(0 , z ) = V (1 - z z 2 + a 2 ) = V ˆ 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 γ the potential on the cylinder is V . We prove the claim by finding a unique solution for x and r . l r -l R r1 r2 g X x (for case V < 0 ) l x (for case V > 0 ) l Figure 1: Equipotential surfaces of two parallel line charges of equal magnitude and opposite polarity. First we note that for symmetry the axis of the cylinder can only be located on a straight line through the two line charges. Since the potential at a distance ρ from a line charge λ is - λ 2 π² 0 ln( ρ ), at the location identified in the figure by γ and r the potential is Φ = λ 2 π² 0 ln

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