《计算电磁学》第九讲ppt课件.ppt
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1、计算电磁学Part II: 矩量法,Dr. Ping DU (杜平),School of Electronic Science and Applied Physics, Hefei University of Technology,E-mail: ,Chapter 2 Electrostatic Fields (静电场),Dec. 5 , 2011,2,Outline,2.1 Operator Formulation (算子描述), 2.2 Charged Conducting Plate (含电荷的导电平板),3,2.1 Operator Formulation,The static ele
2、ctric intensity E is conveniently found from an electrostatic,potential , which is,(2-1),where denotes the gradient operator.,In a region of constant permittivity and volume charge density ,the electrostatic potential satisfies,(2-2),is the Laplacian operator (拉普拉斯算子).,4,For unique solution, the bou
3、ndary conditions on are needed.,In other words, the domain of the operator must be specified.,For now, consider fields from charges in unbounded space, in which case,constant as,(2-3),where r is the distance from the coordinate origin(坐标原点), for every of,finite extent.,The differential operator form
4、ulation is,(2-4),where,(2-5),5,The domain of L is those functions whose Laplacian exists and have bounded at infinity according to (2-3).,The solution to this problem is,(2-6),where is the distance between the source point,( ) and the field point ( ).,Hence, the inverse operator to L is,(2-7),Note t
5、hat (2-7) is inverse to (2-5) only for boundary conditions (2-3).,If the boundary conditions are changed, changes.,6,A suitable inner product for electrostatic problems ( constant) is,That (2-8) satisfies the required postulates (1-2), (1-3) and (1-4) is easily verified.,(2-8),where the integration
6、is over all space.,Let us analyze the properties of the operator L.,For this, form the left side of (1-5),(2-9),where,7,Greens identity is,(2-10),where S is the surface bounding the volume V and n is outward direction normal to S.,Let S be a sphere of radius r, so that in the limit the volume V incl
7、udes all space.,For and satisfying boundary conditions (2-3), and,as .,Hence as . Similarly for .,Since increases only as , the right side of (2-10) vanishes as,. Equation (2-10) then reduces to,(2-11),8,It is evident that the adjoint operator is,(2-12),Since the domain of is that of L, the operator
8、 L is self-adjoint (自伴的).,The mathematical concept of self-adjointness in this case is related to the physical concept of reciprocity.,It is evident from (2-5) and (2-7) that L and are real operators.,They are also positive definite because they satisfy (1-6). For L, form,(2-13),and use the vector i
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