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PHYS809_11F_second_exam_solutions

# PHYS809_11F_second_exam_solutions - November 9th 11th 2011...

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PHYS809 Fall 2011 Second Take-home Exam November 9 th – 11 th , 2011 Please write your name on each answer sheet. Since this is a take home exam, there are a few rules that you must follow. You cannot discuss or communicate about the exam with any one by any means. You cannot look for or use solutions to the problems on the internet. You cannot use printed solutions to Jackson problems. You can use your textbook, notes, and the online course notes. You can use Mathematica or similar software, but you must say when you have used it. The problems do not have equal weight. 1. A ring of radius a has a uniformly distributed total charge Q . The ring is in a concentric spherical cavity of radius b (> a ) . Outside the cavity is dielectric of relative permittivity . r ε (a) Find the potential and electric field everywhere as expansions in Legendre polynomials. (b) Find the surface polarization charge density induced on the surface of the cavity. What is the total surface polarization charge? (c) The ring is replaced by a uniformly charged disk of radius a and total charge Q . Use your solution to part (a) to find the potential everywhere. Find the surface polarization charge density induced on the surface of the cavity. What is the total surface polarization charge? Compare with the result in part (b). (a) We can find the potential of the ring in the absence of dielectric by first finding the potential on the axis of symmetry. This is ( ) 2 2 0 1 . 4 Q z z a πε Φ = + Series expansions are ( ) 2 2 1 0 0 1 , 2 4 n n n z Q z z n πε < + = > - Φ = where ( ) ( ) min , , max , . z z a z z a < > = = Inserting the Legendre polynomials, we find the potential everywhere is ( ) ( ) 2 2 2 1 0 0 1 , cos , 2 4 n n n n r Q r P r n θ θ πε < + = > - Φ = where ( ) ( ) min , , max , . r r a r r a < > = = To take into account the dielectric, we add a potential of form ( ) ( ) 2 1 2 2 1 0 0 , cos , 4 n n n n n r Q r P r θ α θ πε < + = > Φ =

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where ( ) ( ) min , , max , . r r b r r b < > = = Since the normal component of D is continuous at , r b = we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 0 0 2 2 2 2 2 2 0 0 1 1 2 1 cos 2 cos 2 1 1 2 1 cos 2 1 cos . 2 n n n n n n n n r n n n n n n a n P n P b b n a n P n P b b n θ α θ ε θ α θ + = = + = = - - + + = - - + - + This leads to ( )( ) ( ) 2 2 1 2 1 1 .
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