第三课电磁场与电磁波英文版课件.ppt

1、3 Boundary Value3 Boundary Value Problems Problems3 Boundary Value Problems(1)Distribution problems(2)Boundary value problems1.The types of problems in electrostatic field2.The calculation methods of boundary value problems(1)Analytical method(2)Numerical methodBoundary conditionsEquations2V02211212

2、SnnordVdqV3 Boundary Value Problems3.1 Types of Boundary Conditions and Uniqueness Theorem3 Boundary Value Problems Consider a region V bounded by a surface S.1.Dirichlet boundary conditions specify the potential function on the boundary.where f is a continuous function.2.Neumann boundary conditions

3、 specify the normal derivative of the potential function on the boundary.where g is a continuous function.3.Mixed boundary conditions,Sf,Sgn211212SnnThe types of boundary conditions3 Boundary Value Problems The Uniqueness Theorem The uniqueness theorem states that there is only one(unique)solution t

4、o Poissons or Laplaces equation satisfied given boundary conditions.,SfPoissons equation02Laplaces equation,SgnBoundary conditions2V3 Boundary Value Problems Prove(proof by contradiction)Consider a volume V bounded by some surface S.Suppose that we are given the charge density throughout V and the v

5、alue of the scalar potential on surface S.Assume that there exist two solutions and of Laplaces equation subject to the same boundary conditions.Then,0V123 Boundary Value Problems and Let then Applying Greens first identity We have2102203122302dddVVSuv Vuv Vu vSgg3 Boundary Value Problemswhere volum

6、e V bounded by the enclosed surface S.The integral is equal to zero since on the surface S.Thus,The integral can be zero is if is a constantWe know that on the surface S,so we get 222333333dddVVSVV Sg33d0SSg23d0VV312constant30303303 Boundary Value Problems That is,Throughout V and on surface S.Our i

7、nitial assumption that and are two different solutions of Laplaces equations,satisfying the same boundary conditions,turns out to be incorrect.Hence,there is a unique solution to Laplaces equation satisfied given boundary conditions.12123 Boundary Value Problems3.2 Direct IntegrationNote:The potenti

8、al field is a function of only one variable.3 Boundary Value Problems Solution Since the two conductors of radii a and b form equipotential surfaces,the potential must be a function of only.Example 3.2.1The inner conductor of radius a of a coaxial cable is held at a potential of U while the outer co

9、nductor of radius b is grounded.Determine(a)the potential distribution between the conductors,(b)the surface charge density on the inner conductor,and(c)the capacitance per unit length.3 Boundary Value Problems Thus,Laplaces equation reduces to Integrating twice,we obtain(1)where and are constants o

10、f integration to be determined from the boundary conditions.Substituting and into(1),we have (2)Substituting and into(1),we have12lnCC1C2C,b021lnCCb,aU12lnUCaC(3)3 Boundary Value ProblemsHence,the potential distribution within the conductors isThe electric field intensity is1lnUCabablnlnbUablnUba Ee

11、eThus,2lnlnUbCab The normal component of D at is equal to the surface charge density on the inner conductor.Thus,anee3 Boundary Value Problems The charge per unit length on the inner conductor is0lnSUbaa02dlnSSUqSbaFinally,we obtain the capacitance per unit length as02lnqCbUa0SnnDE12adSSqS3 Boundary

12、 Value ProblemsproblemThe inner conductor of radius a of a coaxial cable is held at a potential of U while the outer conductor of radius b is grounded.The space between the conductors is filled with two concentric layers of dielectric.Determine(a)the potential distribution between the conductors and

13、(b)the electric field intensity between the conductors.12c3 Boundary Value ProblemsExample 3.2.2A spherical capacitor is formed by two concentric spherical shells of radii a and b,as shown in Figure 3.2.2.The region between the inner and outer spherical conductors is filled with a dielectric of perm

14、ittivity If U is the potential difference between the two conductors,determine(a)the potential distribution between the conductors,and(b)the capacitance of the spherical capacitor.3 Boundary Value Problems Solution Since the two conductors of radii a and b form equipotential surfaces,the potential m

15、ust be a function of only.Thus,Laplaces equation reduces to r(arb)Integrating twice,we obtainwhere C1 and C2 are constants of integration to be determined from the boundary conditions.121CCr(1)3 Boundary Value Problems Substituting boundary conditions and into(1),we obtain The normal component of D

16、at yields the surface charge density on the inner conductor.Thus,rb0,raUabUaUba rbara000Srrr abUDEra ba 244SqaabCUUbaHence,the capacitance of the system is 3 Boundary Value Problems0)dd(dd122222rrrr)(ra通解2031122401()()6rCrCCrCrr例 体电荷均匀分布在一球形区域内，求电位及电场。解：采用球坐标系,分区域建立方程边界条件arar21ararrr2010有限值01 r参考电位0

17、2r2201120d1 d()ddrrrr)(ar图1.4.2 体电荷分布的球体 0123 Boundary Value Problems电场强度（球坐标梯度公式）：11)(rE3022220()3rraarrr E ree得到220103020()(3)06()3rarraararr 图 随r变化曲线E，errrsin11eer01003rrrraree03 Boundary Value Problems3.3 Separation of VariablesSolve Laplaces equationTransform the three-dimensional partial diffe

18、rential equation into three one-dimensional ordinary differential equations.203 Boundary Value Problems 1.Separation of Variables for Rectangular Coordinate System Laplaces equation is (1)Assume that the potential can be written as the product of one-dimensional potentials.(2)22222220 xyz(,)()()()x

19、y zf x g y h z where is a function of x only,is a function of y only and is a function of z only.()f x()g y()h zSubstituting(2)into(1),and then dividing by we obtain()()(),f x g y h z22222220 xyz(,)()()()x y zf x g y h z3 Boundary Value Problems (3)Since each term involves a single variable,the equa

20、tion is right if and only if each term must be a constant.Thus,we let 2222221d()1d()1d()0()d()d()df xg yh zf xxg yyh zz222d()()dxf xk f xx 222d()()dyg yk g yy 222d()()dzh zk h zz where ,and are called the separation constants.xkykzk2220 xyzkkkNote:The form of general solution to each equation depend

21、s on the separation constant.3 Boundary Value Problems The form of solution of as follows.1.If is a real number,the general solution is 2.If is a imaginary number the general solution is 3.If kx is zero,then 12()exp()exp()xxf xCxCxxk12()sin()cos()xxf xAk xAk xxkj,x12()sh()ch()xxf xBxBx12()f xD xDor2

22、22d()()dxf xk f xx where A1,A2,B1,B2,C1,C2,D1 and D2 are arbitrary constants.hyperbolic sinehyperbolic cosinee-esh()2xxxe+ech()2xxx3 Boundary Value Problems Example 3.3.1 An infinitely long rectangular trough is formed by four conducting planes,locate at and a and and b in air,as shown in Figure 3.3

23、.1.The surface at is at potential of U,the other three are at zero potential.Determine the potential distribution inside the rectangular trough.0 x 0y ybTwo-dimensional field3 Boundary Value Problems Solution Since the trough is infinitely long in the z-direction,the potential can only depend on x a

24、nd y.Laplaces equation reduces to (1)Let (2)Substituting(2)in(1)and dividing by(2),we get 222220 xy(,)()()x yf x g y22221d()1d()0()d()df xg yf xxg yy222222d()()dd()()dxyf xk f xxg yk g yy Let Then,we have220 xykk22xykk or3 Boundary Value Problems Solution Since the trough is infinitely long in the z

25、-direction,the potential can only depend on x and y.Laplaces equation reduces to Using separation of variables,we let Then,we have or Since the boundary conditions are at and we can assume and The form of general solution of the potential is 222220(0,0)xaybxy220 xykk22xykk 00 x,xb20 xk 20yk 1212(,)(

26、shch)(sincos)xxxxx yBk yBk yAk xAk x(,)()().x yf x g ywhere A1,A2,B1 and B2 are arbitrary constants.3 Boundary Value ProblemsThe condition for all requires thenThe condition for all requiresthenThe condition for all requires(0,)0y0yb20,A 121(,)(shch)sinxxxx yBk yBk y Ak x(,)0a y0yb(1,2,3,)xnkna(,0)0

27、 x0 xa20.B Thus,we assume that the general solution is1(,)shsinnnnnx yCyxaaThe condition for all requires(,)x bU0 xa1212(,)(shch)(sincos)xxxxx yBk yBk yAk xAk x11(,)shsinxxx yB Ak yk x3 Boundary Value ProblemsLet This is a Fourier sine series.The coefficients are determined by 1shsinnnnnUCbxaa1sinnn

28、nUbxanb02sind(1,2,3,)ananbUx xna2cos()-1Unn 3 Boundary Value ProblemsWe can get Thus,the general solution of the potential is02,4,(even)41,3,(odd)nnbUnn41,3,shnUCnn bna1,3,5,sinsh4(,)shnnnxyUaax yn bna03 Boundary Value Problems3 Boundary Value Problems Example 3.3.2 Two semi-infinite conducting plat

29、es separated d are grounded,as shown in Figure.A conducting plate between two plates is held at a potential U0.Find the potential between the two conducting plates.Solution The potential can only depend on x and y.Laplaces equation reduces to222220 xy(1)3 Boundary Value Problems Let (2)Substituting(

30、2)in(1)and dividing by(2),we get (,)()()x yf x g y22221d()1d()0()d()df xg yf xxg yy222222d()()dd()()df xf xxg yg yykkLet 1212(ee)(cossin)kxkxCCAkyAkyThe form of general solution of the potential is 3 Boundary Value Problems Using separation of variables,we can write the form of the solution as where

31、 and separation constant k are determined by the boundary conditions.1212(ee)(cossin)kxkxCCAkyAky1212,C CA ASolution The potential can only depend on x and y.Laplaces equation reduces to222220 xy(1)3 Boundary Value ProblemsThe condition for all requires The condition for all requires Since the poten

32、tial is equal to zero at and we obtain The form of the general solution reduces to The condition for all requires(,0)0 x0 x 10.A(,)0 x b0 x 1,2,3,nknbL0yb,x 10C 0(0,)yU0yb1212(ee)(cossin)kxkxCCAkyAky3 Boundary Value Problems This is a Fourier sine series.The coefficients are determined by Thus,the g

33、eneral solution of the potential is01sinnnn yUbDnD2b3 Boundary Value Problems 2.Separation of Variables for Cylindrical Coordinate System (two-dimensional field)Assume that the potential is the function with respect to and The Laplaces equation is (2).(1)(,)()()fg Let(4)()(),fgSubstituting(2)into(1)

34、,we obtainand then multiplying by we obtain(3)()3 Boundary Value ProblemsIn many problems,the potential is a function with respect to with a period .That is (7)Thus,k must be an integer.Taking (6)becomes (8)Each term must be a constant.Thus,we let (5a)(5b)Solving(5b),we have (6)2221d()()dgkg()sincos

35、gCkDk()(2),kn()sincosgCnDn222d()()0dgk g2sincossin(2)cos(2)CkDkCkkDkk3 Boundary Value Problems Substituting n for k,the equation(5a)becomes (9)This is an Euler equation and the solution is (10)Thus,the general solution can be written as 2222d()d()()0ddffn f()nnfAB1(,)()(sincos)nnnnnnnABCnDn n23 Boun

36、dary Value Problems Example 3.3.3 A very long dielectric cylinder of radius a is along z axis in a uniform electric field which is in the direction of x axis.Determine the potential and the electric field intensity.()0E123 Boundary Value ProblemsSolution The form of the solution of the Laplaces equa

37、tion isThe boundary conditions are as follows.(1)when (2)when is finite.(3)when (4)when 21(,)()(sincos)nnnnnnnABCnDn,200cosE xE 0,1,a,a120aa1211(,)()(sincos)nnnnnnnABCnDn (1)(2)121(,)()(sincos)nnnnnnnABCnDn 3 Boundary Value ProblemsFrom boundary condition(1),we haveThus,Let thenBoundary condition(2)

38、requires Thus,Using condition(3),we obtain 01()(sincos)cosnnnnnnnABCnDnE 0,nC and0(1),nDn110ADE 11,B DG201coscosEG 0.nB 11110coscoscosB DADE 11(sincos)nnnnnACnDn(3)201coscosEG 3 Boundary Value ProblemsThus,Let we haveThen,the potential is From condition(4),we have 011(sincos)coscosnnnnnA a CnDnE aCa

39、 0(1)nDn 0,nC 11,A DG1cosG01G aE aGa 00021GEGa 11101coscoscosADaE aGa (4)(6)(5)201coscosEG 3 Boundary Value ProblemsSolving(5)and(6),we obtainThus,the potentials areUsing0002GE 2000Ga E01002cos()Ea 2020001coscos()Ea Ea 1 Eee1cosG201coscosEG 3 Boundary Value Problems000010022(cossin)xEEEeeeWe obtain

40、E2E0ex3 Boundary Value Problems图 均匀外电场中介质圆柱内外的电场 3 Boundary Value Problems 3.Separation of Variables for Spherical Coordinate System (two-dimensional field)Assume that the potential is the function with respective to r and The Laplaces equation reduces to (1)Let (2).(,)()()rf r gSubstituting(2)into(

41、1),we obtain (3)()gr2 r2f(r)3 Boundary Value Problems Multiplying by we obtain (4)Let (5a)(5b)Let (5b)becomes2()(),rf r gcos.u (6)This is the generalized Legendre equation.（勒让德方程）3 Boundary Value Problems Take (7)The solutions are Legendre polynomials (8)(1)0,1,2,kn nn().nP u21d()(1)2!dnnnnnP uunu21

42、d(cos)(cos1)(0)2!d(cos)nnnnnPn3 Boundary Value ProblemsThe former terms of the are (9a)(9b)(9c)(9d)(9e)(9f)(cos)nP0(cos)1P1(cos)cosP221(cos)(3cos1)2P331(cos)(5cos3cos)2P4241(cos)(35cos30cos3)8P5351(cos)(63cos70cos15cos)8P3 Boundary Value Problems3 Boundary Value Problems Consider(5a).(10)The solutio

43、n is (11)Thus,the potential is (12)Constants and are determined by the boundary conditions.(1)()nnnnf rA rB r(1)0(,)(cos)nnnnnnrA rB rPnAnB222()d()21()0ddd f rf rrrn nf rrr3 Boundary Value Problems Example 3.3.4 A dielectric sphere of radius a is in a uniform electric field which is in the direction

44、 of z axis,as in shown in Figure 3.3.5.Determine the potential and the electric field intensity.()0E123 Boundary Value Problems Solution The form of the solution of the Laplaces equation is and Use boundary conditions to determine constants.（1）when The expressions of the fields have only the first t

45、erm That is,(1)10(,)(cos)()nnnnnnrA rB rPra(1)20(,)(cos)()nnnnnnrC rD rPra,r 20001(,)cos(cos)rE zE rE rP ，(1).n 2201(,)cosrE rDr（1）（2）（3）C1=-E0(1)0(,)(cos)nnnnnnrA rB rP2201(,)cosrE rDr 3 Boundary Value Problems(2)when is finite value.So(3)when (5)(4)when we obtain (7)Solving(5)and(7),we obtain0,r 1

46、(,)r0nB,ra12r ar a2101coscosAaE aDa,ra120,r ar arr310012coscosAEDa10(,)(cos)nnnnrA r Pn=111(,)cosrAr010032AE 301002Da E（4）（6）andThe potential is(1)10(,)(cos)()nnnnnnrA rB rPra2201(,)cosrE rDr 3 Boundary Value Problems Thus,the potentials are Using we obtain 01003(,)cos()2rE rra 3020020cos(,)cos()2rE

47、 ra Erar 0001000000333cossin222rzEEEEeee,E3 Boundary Value Problems图图 均匀场中放进了介质球的电场均匀场中放进了介质球的电场E0ez3 Boundary Value ProblemsA conducting sphere of radius a is in a uniform electric field which is in the direction of z axis,as in shown in Figure.Determine the potential and the electric field intensi

48、ty.3 Boundary Value Problems3.4 Method of Images3 Boundary Value Problems Example 3.4.1 Suppose that a point charge q is located above the surface of an infinite conducting plane and grounded,as shown in Figure.Determine(a)the electric potential and electric field intensity above the plane,and(b)the

49、 total charge induced on the surface of the conducting plane.yz3 Boundary Value ProblemsAn electric dipole：(1)The potential at any point on the bisecting plane is zero,(2)The electric field intensity is normal to the plane.yz0The imaginary charge q is said to be the image of the real charge+q3 Bound

50、ary Value Problems Solution(method of images)The boundary condition is .Place an imaginary charge q at(0,0,-d)and temporarily ignore the existence of the plane.The potential at any point P above the plane is(,0)0 x y(zero potential at infinity)The electric field intensity is3 Boundary Value Problems