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1、3 Boundary Value Problems,(1)Distribution problems,(2)Boundary value problems,1.The types of problems in electrostatic field,2.The calculation methods of boundary value problems,(1)Analytical method,(2)Numerical method,Boundary conditions,Equations,or,dq,3.1 Types of Boundary Conditions and Uniquene
2、ss Theorem,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 specify the normal derivative of the potential function on the boundary.where g is a continuous function
3、.3.Mixed boundary conditions,The types of boundary conditions,The Uniqueness Theorem The uniqueness theorem states that there is only one(unique)solution to Poissons or Laplaces equation satisfied given boundary conditions.,Poissons equation,Laplaces equation,Boundary conditions,Prove(proof by contr
4、adiction)Consider a volume V bounded by some surface S.Suppose that we are given the charge density throughout V and the value of the scalar potential on surface S.Assume that there exist two solutions and of Laplaces equation subject to the same boundary conditions.Then,and Let then Applying Greens
5、 first identity We have,where volume 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,That is,Throughout V and on surface S.Our initial assumption that and are two differe
6、nt 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.,3.2 Direct Integration,Note:The potential field is a function of only one variable.,Solution Since the two c
7、onductors 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 conductor of radius b is grounded.Determine(a)the potential distribution between the conductors
8、,(b)the surface charge density on the inner conductor,and(c)the capacitance per unit length.,Thus,Laplaces equation reduces to Integrating twice,we obtain(1)where and are constants of integration to be determined from the boundary conditions.Substituting and into(1),we have(2)Substituting and into(1
9、),we have,(3),Hence,the potential distribution within the conductors isThe electric field intensity is,Thus,The normal component of D at is equal to the surface charge density on the inner conductor.Thus,The charge per unit length on the inner conductor is,Finally,we obtain the capacitance per unit
10、length as,1,problemThe 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(b)
11、the electric field intensity between the conductors.,Example 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 permittivity If U is the potential
12、 difference between the two conductors,determine(a)the potential distribution between the conductors,and(b)the capacitance of the spherical capacitor.,Solution Since the two conductors of radii a and b form equipotential surfaces,the potential must be a function of only.Thus,Laplaces equation reduce
13、s to,(arb),Integrating twice,we obtainwhere C1 and C2 are constants of integration to be determined from the boundary conditions.,(1),Substituting boundary conditions and into(1),we obtain The normal component of D at yields the surface charge density on the inner conductor.Thus,Hence,the capacitanc
14、e of the system is,通解,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,参考电位,图1.4.2 体电荷分布的球体,0,电场强度（球坐标梯度公式）：,得到,图 随r变化曲线,0,3.3 Separation of Variables,Solve Laplaces equation,Transform the three-dimensional partial differential equation into three one-dimensional ordinary differential equations.,1.Sep
15、aration 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),where is a function of x only,is a function of y only and is a function of z only.,Substituting(2)into(1),and then dividing by we ob
16、tain,(3)Since each term involves a single variable,the equation is right if and only if each term must be a constant.Thus,we let,where,and are called the separation constants.,Note:The form of general solution to each equation depends on the separation constant.,The form of solution of as follows.1.
17、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,or,where A1,A2,B1,B2,C1,C2,D1 and D2 are arbitrary constants.,hyperbolic sine,hyperbolic cosine,Example 3.3.1 An infinitely long rectangular trough is formed by four conducting planes,
18、locate at and a and and b in air,as shown in Figure 3.3.1.The surface at is at potential of U,the other three are at zero potential.Determine the potential distribution inside the rectangular trough.,Two-dimensional field,Solution Since the trough is infinitely long in the z-direction,the potential
19、can only depend on x and y.Laplaces equation reduces to(1)Let(2)Substituting(2)in(1)and dividing by(2),we get,Let,Then,we have,or,Solution Since the trough is infinitely long in the z-direction,the potential can only depend on x and y.Laplaces equation reduces to Using separation of variables,we let
20、 Then,we have or Since the boundary conditions are at and we can assume and The form of general solution of the potential is,where A1,A2,B1 and B2 are arbitrary constants.,The condition for all requires thenThe condition for all requiresthenThe condition for all requires,Thus,we assume that the gene
21、ral solution is,The condition for all requires,Let This is a Fourier sine series.The coefficients are determined by,We can get Thus,the general solution of the potential is,Example 3.3.2 Two semi-infinite conducting plates separated d are grounded,as shown in Figure.A conducting plate between two pl
22、ates 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 to,(1),Let(2)Substituting(2)in(1)and dividing by(2),we get,Let,The form of general solution of the potential is,Using separation of variabl
23、es,we can write the form of the solution as where and separation constant k are determined by the boundary conditions.,Solution The potential can only depend on x and y.Laplaces equation reduces to,(1),The condition for all requires The condition for all requires Since the potential is equal to zero
24、 at and we obtain The form of the general solution reduces to The condition for all requires,This is a Fourier sine series.The coefficients are determined by Thus,the general solution of the potential is,2.Separation of Variables for Cylindrical Coordinate System(two-dimensional field)Assume that th
25、e potential is the function with respect to and The Laplaces equation is(2),(1),Let,(4),Substituting(2)into(1),we obtainand then multiplying by we obtain,(3),In 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 m
26、ust be a constant.Thus,we let(5a)(5b)Solving(5b),we have(6),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,n2,Example 3.3.3 A very long dielectric cylinder of radius a is along z axis in a uniform electric f
27、ield which is in the direction of x axis.Determine the potential and the electric field intensity.,Solution The form of the solution of the Laplaces equation isThe boundary conditions are as follows.(1)when(2)when is finite.(3)when(4)when,(1),(2),From boundary condition(1),we haveThus,Let thenBounda
28、ry condition(2)requires Thus,Using condition(3),we obtain,(3),Thus,Let we haveThen,the potential is From condition(4),we have,(4),(6),(5),Solving(5)and(6),we obtainThus,the potentials areUsing,We obtain,E2,E0,ex,图 均匀外电场中介质圆柱内外的电场,3.Separation of Variables for Spherical Coordinate System(two-dimensio
29、nal field)Assume that the potential is the function with respective to r and The Laplaces equation reduces to(1)Let(2),Substituting(2)into(1),we obtain(3),r2,r2,f(r),Multiplying by we obtain(4)Let(5a)(5b)Let(5b)becomes,(6)This is the generalized Legendre equation.（勒让德方程）,Take(7)The solutions are Leg
30、endre polynomials(8),The former terms of the are(9a)(9b)(9c)(9d)(9e)(9f),Consider(5a).(10)The solution is(11)Thus,the potential is(12)Constants and are determined by the boundary conditions.,Example 3.3.4 A dielectric sphere of radius a is in a uniform electric field which is in the direction of z a
31、xis,as in shown in Figure 3.3.5.Determine the potential and the electric field intensity.,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 term That is,（1）,（2）,（3）,C1=-E0,(2)when
32、 is finite value.So(3)when(5)(4)when we obtain(7)Solving(5)and(7),we obtain,n=1,（4）,（6）,and,The potential is,Thus,the potentials are Using we obtain,图 均匀场中放进了介质球的电场,E0,ez,A conducting sphere of radius a is in a uniform electric field which is in the direction of z axis,as in shown in Figure.Determin
33、e the potential and the electric field intensity.,3.4 Method of Images,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 to
34、tal charge induced on the surface of the conducting plane.,An electric dipole：(1)The potential at any point on the bisecting plane is zero,(2)The electric field intensity is normal to the plane.,The imaginary charge q is said to be the image of the real charge+q,Solution(method of images)The boundar
35、y 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,(zero potential at infinity),The electric field intensity is,where and The normal component of the D field must be equal to the surface charge densit
36、y on the surface of the conductor.,+d,-d,Thus,the total charge induced on the surface is,Basic principle(1)Image charges replace the total charge induced.(2)Maintain the field equation and original boundary conditions.(3)Use the real and image charges to determine the field.Note：(1)Image charges are
37、 located outside the field region.(remain field equation)(2)Determine the number,magnitude and position of the image charges.,The number of image charges is,Example 3.4.2 A point charge q is located at a distance d from the center of a grounded conducting sphere of radius a,as shown in Figure 3.4.2.
38、Compute the surface charge density on the surface of the conducting sphere.,Solution(method of images)The boundary condition is Place an imaginary charge at(0,0,d)within the sphere.Now we temporarily ignore the existence of the conducting sphere.The potential outside the sphere is,From the boundary
39、conditions,we obtain Squaring both sides This is an identity with respect to.Thus,Solving these equations,we get and,Choose and.Thus,the potential outside the sphere is Since the normal component of the D field must be equal to the surface charge density on the surface of the conductor,we have,Thus,
40、the total charge induced on the surface is,图 球外的电场分布,（1）A point charge q is placed at a distance d from the center of a conducting sphere of radius a.,图 不接地金属球的镜像,（2）A conducting sphere of radius a has a charge of Q.A point charge q is placed at a distance d from the center of the sphere.,Consider,图
41、 点电荷位于不接地导体球附近的场图,Example 3.4.5 The plane is the interface between two different dielectrics,as shown in Figure(a).A point charge q is located above the interface.Determine the potential in two regions.(a),(b),(c),Solution The boundary conditions are and at the plane Region 1:Place an imaginary char
42、ge at(0,0,-d).We temporarily assume the permittivity of medium 2 is also The potential at any point P in region1 is,Region 2:We superpose an imaginary charge on the charge q.We temporarily assume the permittivity of medium 1 is also The potential at any point P in region 2 is,Any point on the interf
43、ace satisfiesFrom boundary conditions,we haveSolving these equations,we obtain,Thus,the potential at any point in region 1 isThe potential at any point in Region 2 is,Example 3.4.2 A long straight line charge is parallel a long straight grounded wire of radius a,as shown in Figure.Calculate the pote
44、ntial in space.,Solution(method of images)Place an imaginary line charge on the y=0 plane within the wire.is parallel to.Temporarily remove the wire.The potential at any point P is Since the long straight wire of radius a is grounded,the potentials at points A and B are equal to zero.,Thus,we have,L
45、et We obtain andThus,the potential at any point P is,Trial solution试探解,The surface charge density on the surface of the wire is,The charge induced per unit length on the surface is,Method of electric axis,Example 3.4.4 Two parallel cylindrical conductors of radius a separated a distance 2d form tran
46、smission line,as shown in Figure.Calculate the capacitance per unit length of two cylindrical conductors.,Solution Using method of electric axis,the potential at any point outside of the conductors is The potential at A point is The potential at B point is The potential difference is,(C=0 zero refer
47、ence at y axis),The capacitance per unit length of two cylindrical conductors is When we obtain,Two long straight lines with equal and opposite but uniform charge distributions are separated by a distance of 2b.Find the potential in free space.,(C=0,zero reference at y axis),Let,A set of eccentric c
48、ircles,Thus,we obtain,The equation of equipotential lines is,center,Radius,When the radius is a,circle,根据，得到 Ex 和 Ey 分量,Lines of forces of E,例 不同半径两平行长直导线（传输线）相距为d,确定电轴位置，单位长度的电容。,图 不同半径传输线的电轴位置,解：,镜像法（电轴法）的理论基础是：,镜像法（电轴法）的实质是：,镜像法（电轴法）的关键是：,镜像电荷（电轴）只能放在待求场域以外的区 域。叠加时，要注意场的适用区域。,用虚设的镜像电荷（电轴）替代未知电荷的分
49、布，使计算场域为无限大均匀媒质；,静电场惟一性定理；,确定镜像电荷（电轴）的个数、大小及位置；,应用镜像法（电轴法）解题时，注意：,3.5 The Finite-Difference Method(FDM),Numerical methods:finite-difference method(FDM),finite element method(FEM),method of moments(MOM).The finite-difference method(FDM),基本思想：将场域离散为许多网格，应用差分原理，将求解连续函数 的微分方程问题转换为求解网格节点上 的代数方程组的问题。,Divi
50、de the solution domain into finite discrete points and replace the partial differential equation with a set of difference equations.,If the potential variation is independent of z,Poissons equation simplifies to(1)First,we divide the region into a finite number of meshes.,Let us consider a square me
