This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 nonzero. 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 nonzero only if k is odd. Let k = 2 n +1. The nonzero 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 3D 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 2dimensional free space Green function. In polar coordinates, it is ( ) ( ) 2 2 , , ,...
View
Full
Document
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.
 Fall '11
 MacDonald
 Work

Click to edit the document details