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# Homework_4 - Homework 4 solutions 2.14 (a) Noting that the...

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Unformatted text preview: Homework 4 solutions 2.14 (a) Noting that the boundary conditions are an odd function of , the series solution for the potential inside the cylinder is of form ( ) 1 , sin . m m m a m b = = The coefficients are ( ) 2 3 2 2 2 2 3 2 1 1 1 1 1 , sin sin sin sin sin . m a b m d V m d V m d V m d V m d = =- +- This gives [ ] [ ] [ ] [ ] ( ) 2 3 2 2 2 3 2 cos cos cos cos 2 3 cos 1 cos cos 2 2 2 cos 1 cos cos cos 2 2 2 1 cos 1 1 . 2 m m V a m m m m m V m m m m V m m m m m V m m = -- +- = -- - + = -- - + =- + - We see that only even coefficients are non-zero. Let m = 2 k , so that ( ) ( ) 2 2 2 1 cos 1 1 . k k V V a k k k =- =- - We now see that the coefficients are non-zero only if k is odd. Let k = 2 n +1. The non-zero coefficients are ( ) 4 2 4 , 2 1 n V a n + = + and the series solution is ( ) ( ) ( ) 4 2 4 1 , sin 4 2 . 2 1 n n V n n b + = = + + (b) To sum the series consider ( ) 2 1 2 2 2 1 . 2 1 n i n S e n b + = = + Let 2 2 2 . i z e b = Then 2 2 1 1 1 1 . 1 2 1 1 n n dS z dz z z z = = = = + -- + Hence 1 1 ln . 2 1 z S z + =- Thus ( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 1 4 4 1 1 4 2 1 2 , Im Imln Imln 1 1 1 2 2 Imln 1 tan 1 2 2 sin 2 tan . z z V V z V S z z z V V z z zz z z zz V b b * * * * *- *- +- + = = =---- =- + - =- =- (c) The equipotentials look like: The field lines are perpendicular to the equipotentials and the direction of the field is from high to low potential. 2.17 (a) In 3-D the free space Green function is ( ) ( ) ( ) ( ) 1 2 2 2 2 1 , . G x x y y z z = - +- +- r r Integrating with respect to z z - from Z to + Z , we get ( ) ( ) ( ) 1 2 1 2 2 2 2 2 , , , 2 , Z Z Z du du g x y x y a u a u- = = + + where ( ) ( ) 2 2 2 . a x x y y =- +- Making the substitution sinh , u a v = this becomes ( ) ( ) 2 2 1 , , , 2sinh 2ln . Z Z a Z u g x y x y a a- + + = = Now assume that , Z a so that ( ) 2 2 2 2 1 , , , 2ln 2ln 2ln . Z a Z Z a g x y x y Z a a Z + + = - +- Dropping the large but inessential constant 2ln , Z the leading term in the dependence on a is ( ) ( ) ( ) 2 2 , , , 2ln ln . g x y x y a x x y y = - = -- +- This is what we identify as the 2-dimensional free space Green function. In polar coordinates, it is ( ) ( ) 2 2 , , ,...
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## This note was uploaded on 12/02/2011 for the course PHYS 809 taught by Professor Macdonald during the Fall '11 term at University of Delaware.

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Homework_4 - Homework 4 solutions 2.14 (a) Noting that the...

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