《计算电磁学》第九讲ppt课件.ppt

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 TechnologyE-mail: Chapter 2 Electrostatic Fields (静电场) Dec. 5 , 20112Outline2.1 Operator Formulation (算子描述) 2.2 Charged Conducting Plate (含电荷的导电平板) 32.1 Operator Formulation The static electric intensity E is conveniently found from an electrostaticpotential , which is E(2-1)where denotes the gradient operator. In a region of constant permittivity and volume charge density , the electrostatic potential satisfies2 (2-2) is the Laplacian operator (拉普拉斯算子). 24For unique solution, the boundary 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 rrr (2-3)where r is the distance from the coordinate origin(坐标原点), for every offinite extent.The differential operator formulation is L(2-4)where2L (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 ( , )( , , )4x y zx y zdx dy dzR(2-6)where is the distance between the source point 222()()()Rxxyyzz( ) and the field point ( ). , x y z, ,x y zHence, the inverse operator to L is 114Ldx dy dzR(2-7)Note that (2-7) is inverse to (2-5) only for boundary conditions (2-3). If the boundary conditions are changed, changes. 1L6A suitable inner product for electrostatic problems ( constant) isThat (2-8) satisfies the required postulates (1-2), (1-3) and (1-4) is easily verified. ,( , , ) ( , , )x y zx y z dxdydz (2-8)where the integration is over all space.Let us analyze the properties of the operator L.For this, form the left side of (1-5),2,()Ld (2-9)where ddxdydz7Greens identity is 22VSddsnn (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 includes all space. r For and satisfying boundary conditions (2-3), and 1/Cr2Cnras .r Hence as . Similarly for . 3Cnrr nSince increases only as , the right side of (2-10) vanishes as 2sindsrd d 2rr . Equation (2-10) then reduces to 22dd (2-11)8It is evident that the adjoint operator is aL2aLL (2-12)Since the domain of is that of L, the operator L is self-adjoint (自伴的). aLThe 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.1LThey are also positive definite because they satisfy (1-6). For L, form2*,*()Ld (2-13)and use the vector identity plus the divergence2() Theorem (散度定理).9The result is *,VSLdd s(2-14)where S bounds V. Again take S a sphere of radius r. For satisfying (2-3), the last term of (2-14) vanishes as . r Then2*,|Ld(2-15)and, for real and , L is positive definite. In this case, positive0definiteness of L is related to the concept of electrostatic energy (静电能）.10 2.2 Charged Conducting Plate (含电荷的导电平板) Consider a square conducting 2a meter on a side and lying on the plane 0z with center at the origin as shown in Fig. 2-1. Fig. 2-1. Square conducting plate and subsections11Let represent the surface charge density on the plate. Here, we assume that the thickness is zero.( , )x yThe electrostatic potential at any point in space is ( ,)( , , )4aaaax yx y zdxdyR(2-16)where 222()()()RxxyyzzThe boundary condition is (constant) on the plate. VThe integral equation for the problem is 222( ,)4()()()aaaax yVdxdyxxyyzz(2-17)12where , . xayaThe unknown to be determined is the charge density . ( , )x yA parameter of interest is the capacitance of the plate 1( , )aaaaqCdxdyx yVV(2-18)which is a continuous linear functional of .Let us first go through a simple subsection and point-matching solution, and later interpret it in terms of more general concepts. Consider the plate divided into N square subsections, as shown in Fig. 2-1. Define basis functions nm1ons0on all othersnf(2-19)13Thus the charge density can be represented by1( , )Nnnnx yf(2-20)Substituting (2-20) in (2-17), and satisfying the resultant equation at the mid-point of each , we obtain the set of equations (,)mmxyms1NmnnnVl1,2,mN(2-21)where2214()()nnmnxymmldxdyxxyy(2-22)14Note that is the potential at the center of due to a uniform mnlmscharge density of unit amplitude over . nsA solution to the set (2-21) gives the in terms of which the charge density mis approximated by (2-20). The corresponding capacitance of the plate, approximating (2-18), is111NnnmnnnmnCslsV(2-23)To translate the above results into the language of linear spaces and the method of moments (MoM), let 15( , )( , )f x yx y( , )g x yVxaya222( ,)( )4()()()aaaaf x yL fdxdyxxyyzz(2-24),(2-25)(2-26)Then is equivalent to (2-17). ( )L fgA suitable inner product, satisfying (1-2) to (1-4), for which L is self-adjoint, is,( , ) ( , )aaaaf gdxdyf x y g x y(2-27)We choose the functions (2-19) as a subsectional basis. 16The testing functions are defined as() ()mmmwxxyy(2-28)This is the two-dimensional Dirac delta function. The elements of the l matrix (1-25) are those of (2-22), and the g matrix of(1-26) ismVVgV (2-29)The matrix equation equation (1-24) is identical to the set of equations (2-21). 17In terms of the inner product (2-27), the capacitance (2-18) can be written2,CV For numerical results, the of (2-22) must be evaluated. mnlLet denote the side length of each . 22 /baNnsThe potential at the center of due to unit charge density over its own surface isnsnn22122ln(12)0.88144bbbbbbldxdyxy(2-31)(2-30)18The potential at the center of due to unit charge density over unit charge over can be similarly evaluated, but the formula is complicated. msnsFor most purpose, it is sufficiently accurate to treat the charge on as if it were apoint charge, and usens2n224()()nmmnmnmnsblRxxyy(2-23)19Homework: Repeat this example. The side length is assume to be equal t o 1.Analyze the size of the subsectional domain V.S. the accuracy.20212223242526272829