{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
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
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern