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Unformatted text preview: Electromagnetics Workshop Week 10 Solutions Boundary Conditions & Review for Exam Problem 1: Given the potential, 20 cos = for 2 < and ( 29 50 cos 120 cos = for 2 : a) Find E in both regions. b) Show that the boundary conditions are satisfied at the surface 2 = for all if = for 2 and r = for 2 < . Find r . Problem 1 Solution: For cylindrical coordinates: 1 z =  =  + + E z ( 29 ( 29 ( 29 ( 29 1 20cos 20 sin 20cos 20sin , 2 =  =  + < E ( 29 ( 29 ( 29 ( 29 2 2 cos 50 120 sin 50 120 , 2 =  + + E At 2 = , the tangential components of the electric field should be equal: ( 29 ( 29 ( 29 2 2 20sin sin 50 120 = = The tangential electric field components are equal for all at the surface 2 = . At 2 = , the normal components of the electric flux density should be equal: ( 29 ( 29 ( 29 2 2 20cos cos 50 120 r = =  + The normal electric flux density components are equal for all at the surface 2 = if 4 r = ....
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This note was uploaded on 10/25/2011 for the course ECE 2317 taught by Professor Staff during the Spring '08 term at University of Houston.
 Spring '08
 Staff
 Electromagnet

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