F07_Practice_Final_Solutions

F07_Practice_Final_Solutions - Physics 505 Fall 2007...

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Unformatted text preview: Physics 505 Fall 2007 Practice Final — Solutions This final will be a three hour open book, open notes exam. Do all four problems. 1. Two point charges q and- q are located on the z axis at z = + d/ 2 and z =- d/ 2, respectively. a ) If the charges are isolated in space, show that the potential admits a Legendre expansion Φ( r,θ ) = 2 q 4 π X l odd r l < r l +1 > P l (cos θ ) where r < = min( r,d/ 2) and r > = max( r,d/ 2). From basic considerations, the Coulomb potential of the two point charges can be written as Φ = 1 4 π q | ~x- ( d/ 2)ˆ z |- q | ~x + ( d/ 2)ˆ z | We now recall the expansion 1 | ~x- ~x | = X l r l < r l +1 > P l (cos γ ) where γ is the angle between ~x and ~x . Since the two charges are located on the z-axis, the angle γ is simply θ for the positive charge and π- θ for the negative charge. Thus Φ = q 4 π X l r l < r l +1 > [ P l (cos θ )- P l (- cos θ )] Since P l (- ζ ) = (- 1) l P l ( ζ ), the even l components cancel out, and we are left with Φ = 2 q 4 π X l odd r l < r l +1 > P l (cos θ ) b ) Now consider the charges to be contained inside a linear dielectric sphere of permittivity and radius a (where a > d/ 2).- q q Find the electric potential everywhere as an expansion in Legendre polynomials. The dielectric sphere separates all of space into two regions, r < a and r > a . We thus write down expansions for the electrostatic potential inside and outside the sphere, and match at the boundary. In particular, introduce Φ in = 2 q 4 π X l odd r l < r l +1 > + A l r l P l (cos θ ) Φ out = 2 q 4 π X l odd B l r l +1 P l (cos θ ) where r < = min( r,d/ 2) and r > = max( r,d/ 2). Since we match at the surface r = a we use the fact that r < = d/ 2 and r > = r near this surface. The matching conditions are then on D ⊥ and E k . We find D ⊥ :- ( l + 1) ( d/ 2) l a l +2 + lA l a l- 1 =- ( l + 1) B l a l +2 E k : ( d/ 2) l a l +2 + A l a l- 1 = B l a l +2 or, in matrix form la 2 l +1 l + 1- a 2 l +1 / A l B l = ( d/ 2) l l + 1 1 which may be solved to yield A l = ( r- 1)( l + 1) ( r + 1) l + 1 ( d/ 2) l a 2 l +1 , B l = 2 l + 1 ( r + 1) l + 1 ( d/ 2) l where r = / . This gives explicitly Φ in = 2 q 4 π X l odd r l < r l +1 > + ( r- 1)(...
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This note was uploaded on 11/21/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.

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F07_Practice_Final_Solutions - Physics 505 Fall 2007...

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