Chapter 4. Problems of Boundary Values in Static Electric Fields

1. 1) Separation of variables in system of rectangular coordinate 2) Separation of variables in system of cylindricalcoordinate 3) Separation of variables in system of spherical coordinate 4) Image method (镜像法不作要求) Chapter 4. Problems of Boundary Values in Static Electric Fields ◇ The boundary value problem of electrostatic and steady fields can be solved by Laplace or Poission equation under given boundary conditions

2. 回忆两无源介质间(无电荷无电流)的边界条件

3. 1 Separation of variables in system of rectangular coordinate Separation of variables Separation of variables is a classical differential equation method, which is suitable for a sort of boundary value with ideal boundary conditions. ◇ General steps  Select the coordinate according to the geometric shape and field distribution of the boundary, and list the differential equations and boundary conditions.  Separate the variables.  Solve the equation (general solution: the linear combination of all special solutions)  Determine the integral constant (based on the given boundary conditions) to obtain the final solution

4. An example: Determine the potential function in 2D rectangle space. y b According to the problem, the potential function and its boundary conditions are given as follows: x a

5. In rectangular coordinate, can be expressed as: (1) Separate the variables as follows: (2) Substituting equation (2) into (1), one can arrive at (3)

6. (4) Only depends on variable x Only depends on variable y So the two terms on the left side in Eq. (4) can be equal to two constant numbers, respectively, i.e., (5) The two sides of the above equation is divided by

7. Where (6) Thus the solution of the equation (5) is: (7) If k x is a real number and k y an imaginary number Or (8) If k x is an imaginary number (9) If k x=0

8. Analyzing the boundary conditions, we can obtain (7) Since =0 when x=a and 0 Substituting the boundary conditions into Eq. (7), we can get (8) where is called the eigenvalue of the present boundary problem:

9. Thus, g(y) is (9) Note: (10) Hence, The general solution is given by (11)

10. Employing the theory of Fourier series expansion (12) So the final solution is: (13)

11. The Laplace equation in cylindrical coordinate system can be written as: (14) If the field is two-dimensional field and  is independent of z, the Laplace equation can be rewritten (simplified) as (15) Assume the solution (r, ) can take the following form (16) 2 Separation of variables in system of cylindrical coordinate

12. Substitution of (16) into (15) yields (17) Simplify Eq. (17) by multiplying (18) Function of r Function of  So, (19)

13. Then, (20) and we obtain (21) As the potential function should satisfy..., this means ＝n (integer) (22) Therefore, (23)

14. The left side of Eq. (19) can be rewritten as (24) i.e., (25) （Euler equation） The solution to Eq. (25) is (26) The general solution of  in the two-dimensional cylindrical coordinate is (27)

15. Example: an infinitely long dielectric cylinder with radius a and permittivity  is placed in an externaluniform electrical field E0 vertical to cylinder.If the field is directed along x-axis and the axis of cylinder is z-axis, determine the potential functions inside and outside the cylinder. In cylindrical coordinate, x=r cos, The corresponding potential of E0 is: (28) If the potentials inside and outside the cylinder are 1 and 2, respectively, the equations and boundary conditions are given as follows

16. Finite (29) 有限 So, (30) From condition (31) and (32)

17. (33) Because is finite (34) Since , we have (35)

18. Because (36) Thus, (37) The potentials are given by (38) (39)

19. The Laplace equation in spherical coordinate can be written as (40) The potential function can be expressed as: Without considering  (41) 3 Separation of variables in spherical coordinate system

20. Separate variables (in the similar fashion), we can get the general solution to equation (14) (42) where ◇ Legendre polynomial:

21. Example: a spherical dielectric with radius a and permittivity  is placed in a uniform electrical field E0. If the field is along z-axis, determine the potentials inside and outside the sphere. In spherical coordinate system, z=r cos . The corresponding potential of E0 can be expressed as Assume the potentials inside and outside the sphere are 1 and 2, respectively. The equations and boundary conditions are as follows

22. 有限 finite The general solution is Because and

23. is finite Because Since

24. Method of Image • (镜像法不作要求) According to the uniqueness theorem, in some appropriate position outside of the field under consideration, we can utilize some dummy (虚拟的) charges (image charge) to replace the induced charges of the conductor interface or the polarization charges on the media interface. Transform the original problem into the equivalent problem with the same boundary conditions. The principles to select image charges: 1. Image charge should be outside of the region to be investigated(The image charge must be external to the volume of interest) 2. The boundary conditions should not be changed after introducing the image charges. 3. Image method can only be suitable for those special boundaries (such as plane, circular and spherical boundaries).

25. ●镜像法基本思路：在所研究的场域外的某些适当位置，用一些虚拟电荷等效替代导体分界面上的感应电荷或媒质分界面上的极化电荷的影响。●镜像法基本思路：在所研究的场域外的某些适当位置，用一些虚拟电荷等效替代导体分界面上的感应电荷或媒质分界面上的极化电荷的影响。 ●也即：在研究区域之外，用一些假想的电荷分布代替场问题的边界。——将原问题转为与原问题边界条件相同的等效问题。 ●镜像法求解电位问题的理论依据：唯一性定理。待求区域的电位由其电荷分布与边界条件共同决定。 ●等效电荷一般位于原电荷关于边界面的镜像点处，故称为镜像电荷。大多是一些点电荷或线电荷。 ●镜像电荷位置选择原则： 1、镜像电荷必须位于求解区域以外的空间。 2、镜像电荷的引入不能改变原问题的边界条件。

26. 1. plane image Infinite grounding conductor plane (z=0), where a point charge is placed at z=h. Determine the potential distribution of the upper half space. Because the original charge is placed at z=h, its mirror image should be at z=-h with the quantity being -q. Thus the potential at boundary z=0 keeps zero, and the conductor plane at z=0can be removed in the equivalent system.

27. So the potential in z>0 is written as (43) The total induced charges in infinite conductor is (44) The quantity of original and induced charges are equal to each other. (45)

28. 2. Spherical face image A grounding conductor sphere has the radius a. A point charge is placed at P1 point. The distance between charge and spherical center O is d1. Determine the potential distribution outside the sphere. Because the potential outside the sphere is decided by point charge and induced charge on the conductor surface, the latter of which can be replaced by the image charge q2 if the potential keep zero on spherical surface.

29. So, the potential outside the sphere can be written as (46) In order to determine the position and quantity of the image charge, we can consider two special points, say, S1 and S2, inside the sphere, where the potentials are zero, namely, (47)

30. The solutions to the above two equations are given by (48) or given up Thus, the potential outside the sphere can be rewritten as (49) where (50)

31. (以下内容根据课时情况决定增删)Laplace Equations in Spherical and Cylindrical Coordinate Systems球坐标和柱坐标系中的拉普拉斯方程

32. Laplace方程的解（勒让德函数、贝塞尔函数等）称为“特殊函数”。这些解都是确定的、固定的，已经在100多年前被英美数学家求出，主要用了幂级数展开法（power series expansion）. 求解过程是很复杂的，但今天的我们的任务就是利用这些解来表示任何一个边值问题的解，因为任何一个边值问题的解均可以用这些特殊函数来展开。我们的目的就是利用边界条件来定出这些展开系数。 下面的内容来自Jackson经典电动力学“Classical Electrodynamics”(John Wiley & Sons, 1998, pp.95-119) (可见复印的资料). 建议自学。

33. Boundary-Value Problems 1 Laplace’s Equation in Spherical Coordinates In spherical coordinates We can solve this eq. by separation of variables, let

34. Multiplying We have Clearly the solution is In order that Q be single valued , m must be an integer Similarly, we separate and

35. Introducing another real constant For r we have, Let x=cos Its solutions are the associated Legendre functions For m=0, - ----- Legndre Polynomials

36. (Rodrigues formula)‏ Orthogonal: So the general solutions of Lapace eq. in Spherical Coordinates

37. 2 Boundary-value problems with azimuthal Symmetry A problem possessing azimuthal symmetry m=0, then As usual, the coefficients can be determined by the BCs. Suppose Vis the potential on S of sphere of radius a. We want to know the potential V inside the sphere. If no charge at the origin, -- finite, leading to for all l. From the relation we have

38. If, for example, in hemisphere, To find V outside the sphere The series is a unique expansion with BCs. So from a knowledge of V in a limited domain, namely on the symmetry axis, we may derive the solution e.g. z=r Valid for positive z. If this potential function can be expanded in a power series in z=r , with known coefficient, the solution is derived by multiplying by

39. Let us see hemisphere again. We have obtained This can be expanded in powers of By comparison with only , i.e., terms with odd l are non-zero. This is the same as the previous result.

40. An important expansion is that the potential at due to a unit point charge at Where is the smaller (larger) of -- the angle between This can be proved by rotating axis so that , lies alone the z axis. Then the potential satisfies Lapace’s eq., possesses azimuthal symmetry, and can be expanded to . except at point

41. If is on the z axis, the RHS reduces to While the LHS becomes For points off the axis , it is only necessary to multiply each term in the above eq. by The general result is proved.

42. Another example is the potential due to a total charge q uniformly distributed around a circular ring of radius a, located as shown in the Figure, with its axis the z axis and its center at z=b. The potential at a point P on the axis The inverse distance AP, 1/AP can be expanded by using

43. The potential at any point in space is then derived by multiplying each member of the series by the Legendre polynomials Where is the smaller (larger) of r and c

44. 3 Associated Legendre Polynomials and the Spherical Harmonics So far, we have considered the azimathal symmetry m=0, these involve only ordinary Ledgendze polynomials. Legendre’s polynomials

45. The general potential problem can, however, have variations so that in In this case, we need the generalization of It can be shown that in order to have finite solution on the interval the parameterl must be zero or a positive integer and the integer m can take on only positive m (+|m|,-|m|; Rodrigues’formula)‏

46. Since differential eq. for depends only on and m is an integer, and are proportional For fixed m, form an orthogonal set in the index l on the interval The sol. of Laplace eq. was decomposed into a product of factors for form a complete set of orthogonal in the index m on the interval

47. form a similar set in the index l for each m for Therefore, will form a complete orthogonal set on the surface of the unit sphere in the two indices l, m. From the normalization condition, it is clear that suitable normalized function, denoted by called spherical harmonics It can be seen

48. The normalization and orthogonality conditions are The complete relation is

49. Note that, for m=0, which recovers the ordinary Legendre functions. Since the spherical harmonics form a complete orthogonal set of function on the surface of the unit sphere, then an arbitrary function can be expanded in spherical harmonics Where the coefficients

50. A point of interest to us is the form for with The general sol. for a Boundary-value problem in spherical coordinates can be written If V is specified on a spherical surface, the coefficient can be determined by this BC.