EM_scattering_homework_1_solution

Now find the scattered field in any direction ks

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Unformatted text preview: � � 0 � � exp � � jky � � V jkz � dV V 4a �� � � 0 � 2 � ka � � sin � � k2 �2� (1.3) ˆ ˆ In the case ks � � z : ˆ � ˆ � � � � 0 � a 3 , ks � � z � ˆˆ� �� � � 0 � � exp � j � � kz � kz � � r � dV � � a2 � ˆ ˆ �� � � 0 � sin � ka � , ks � � z V � � k (1.4) Problem 2: Repeat this calculation for a dielectric cylinder with the radius a , height h , and dielectric ˆ constant � . Now find the scattered field in any direction ks . Solution Let us again focus on the integral � � exp � j � k s ˆ� � ki � � r � �� p ( r )dV � � � � � 0 � � exp � j � ks � kz � � r � dV � � V (1.5) V In cylindrical coordinates: ks � � k s , � cos �s , k s , � sin �s , ks , z � , r � � r cos � , r sin � , z � (1.6) So that the above integral becomes a 2� h / 2 �� � � 0 � � � � 0 0 �h / 2 � �� � � 0 � Note that exp � jk s , � r cos �� � �s � � j � k s , z � k � z � rdrd� dz � � 2 sin �� k s , z � k � h / 2 � � � k s ,z � k (1.7) a 2� � � exp � jk � 00 s,� r cos �� � �s � � rdrd� � 1 2� 2� � exp � jk � s,� r cos �� � �s �� d� � J 0 � k s ,� r � � (1.8) 0 So that the integral yields �� � � 0 � 4� sin � � k s , z � k � h / 2 � � � ks ,z � k � �...
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