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Unformatted text preview: Homework 6 solutions 3.23. The potential is the solution of ( ) ( ) ( ) 2 2 2 2 2 1 1 . z z q z  + + =  To obtain each of the first two forms of the potential, we make use of the completeness relation ( ) ( ) 1 . 2 im m e  =  = Let ( ) ( ) ( ) 1 , , , . 2 im m m z f z e  = = Substituting this into the Poisson equation gives ( ) ( ) ( ) 2 2 2 2 1 , , , . m z z m q f z z z   + =  (1) To derive the first form for the potential, we use the completeness relation ( ) ( ) 2 2 1 1 2 , m mn m mn n m mn J x J x a a a J x = +  = to write ( ) ( ) ( ) 2 2 1 1 2 , , , , . m mn m mn m mn n m mn J x J x a a f z z g z z a J x = + = This leads to the ordinary differential equation ( ) 2 2 2 2 . mn mn mn d g x q g z z dz a  =  The boundary conditions are mn g = at z = and . z L = This requires that ( ) ( ) , sinh sinh , mn mn mn x x g z z A z L z a a < > = where A is a constant that is determined by the jump condition, . mn mn dg dg q dz dz + =  This gives ( ) ( ) sinh cosh cosh sinh sinh . mn mn mn mn mn mn mn x x x x x A z L z z L z a a a a a x x A L a a q  =  =  Hence ( ) ( ) , sinh sinh , sinh mn mn mn mn mn x x q a g z z z L z x a a x L a < > = The potential is therefore ( ) ( ) ( ) ( ) 2 1 1 , , sinh sinh . sinh im m mn m mn mn mn m n mn mn m mn e J x J x x x q a a z z L z x a a a x J x L a  < > = = + = To obtain the second form for the potential we use the completeness relation ( ) 1 2 sin sin n n z n z z z L L L = = in equation (1). We look for a solution ( ) ( ) 1 2 , , , , sin sin . m mn n n z n z f z z g L L L = = Substituting into equation (1), this leads to ( ) 2 2 2 2 2 1 ....
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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.
 Fall '11
 MacDonald
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