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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 conditionsEquationsordq3 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 specify the normal derivative of the potential function on the boundary.where g is a continuous function.3.Mixed boundary conditionsThe types of boundary conditions3 Boundary Value Problems The Uniqueness Theorem The uniqueness theorem states that there is only one(unique)solution to Poissons or Laplaces equation satisfied given boundary conditions.Poissons equationLaplaces equationBoundary conditions3 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 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,3 Boundary Value Problems and Let then Applying Greens first identity We have3 Boundary Value Problemswhere 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 3 Boundary Value Problems That is,Throughout V and on surface S.Our initial 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.3 Boundary Value Problems3.2 Direct IntegrationNote:The potential 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 conductor 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 of integration to be determined from the boundary conditions.Substituting and into(1),we have (2)Substituting and into(1),we have(3)3 Boundary Value ProblemsHence,the potential distribution within the conductors isThe electric field intensity isThus,The normal component of D at is equal to the surface charge density on the inner conductor.Thus,3 Boundary Value Problems The charge per unit length on the inner conductor isFinally,we obtain the capacitance per unit length as13 Boundary 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(b)the electric field intensity between the conductors.3 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 permittivity 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 must be a function of only.Thus,Laplaces equation reduces to(arb)Integrating twice,we obtainwhere C1 and C2 are constants of integration to be determined from the boundary conditions.(1)3 Boundary Value Problems 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 capacitance of the system is 3 Boundary Value Problems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 03 Boundary Value Problems电场强度（球坐标梯度公式）：得到图 随r变化曲线03 Boundary Value Problems3.3 Separation of VariablesSolve Laplaces equationTransform the three-dimensional partial differential equation into three one-dimensional ordinary differential equations.3 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)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 obtain3 Boundary Value Problems (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.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 orwhere A1,A2,B1,B2,C1,C2,D1 and D2 are arbitrary constants.hyperbolic sinehyperbolic cosine3 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 3.3.1.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 field3 Boundary Value Problems Solution Since the trough is infinitely long in the z-direction,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 Then,we haveor3 Boundary Value Problems 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 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.3 Boundary Value ProblemsThe condition for all requires thenThe condition for all requiresthenThe condition for all requires Thus,we assume that the general solution isThe condition for all requires3 Boundary Value ProblemsLet This is a Fourier sine series.The coefficients are determined by 3 Boundary Value ProblemsWe can get Thus,the general solution of the potential is3 Boundary Value Problems3 Boundary Value Problems Example 3.3.2 Two semi-infinite conducting plates 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 to(1)3 Boundary Value Problems Let (2)Substituting(2)in(1)and dividing by(2),we get Let The 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 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)3 Boundary Value ProblemsThe condition for all requires The condition for all requires Since the potential is equal to zero at and we obtain The form of the general solution reduces to The condition for all requires3 Boundary Value Problems This is a Fourier sine series.The coefficients are determined by Thus,the general solution of the potential is3 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)Let(4)Substituting(2)into(1),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)3 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 n23 Boundary 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.3 Boundary Value ProblemsSolution 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)3 Boundary Value ProblemsFrom boundary condition(1),we haveThus,Let thenBoundary condition(2)requires Thus,Using condition(3),we obtain(3)3 Boundary Value ProblemsThus,Let we haveThen,the potential is From condition(4),we have(4)(6)(5)3 Boundary Value ProblemsSolving(5)and(6),we obtainThus,the potentials areUsing3 Boundary Value ProblemsWe obtain 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)Substituting(2)into(1),we obtain (3)r2 r2f(r)3 Boundary Value Problems Multiplying by we obtain (4)Let (5a)(5b)Let (5b)becomes (6)This is the generalized Legendre equation.（勒让德方程）3 Boundary Value Problems Take (7)The solutions are Legendre polynomials (8)3 Boundary Value ProblemsThe former terms of the are (9a)(9b)(9c)(9d)(9e)(9f)3 Boundary Value Problems3 Boundary Value Problems Consider(5a).(10)The solution is (11)Thus,the potential is (12)Constants and are determined by the boundary conditions.3 Boundary Value Problems Example 3.3.4 A dielectric sphere of radius a is in a uniform electric field which is in the direction of z axis,as in shown in Figure 3.3.5.Determine the potential and the electric field intensity.3 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 term That is,（1）（2）（3）C1=-E03 Boundary Value Problems(2)when is finite value.So(3)when (5)(4)when we obtain (7)Solving(5)and(7),we obtainn=1（4）（6）andThe potential is3 Boundary Value Problems Thus,the potentials are Using we obtain 3 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 intensity.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 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.yzThe imaginary charge q is said to be the image of the real charge+q3 Boundary 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(zero potential at infinity)The electric field intensity is3 Boundary Value Problems where and The normal component of the D field must be equal to the surface charge density on the surface of the conductor.+d-dThus,the total charge induced on the surface is3 Boundary Value Problems 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 located outside the field region.(remain field equation)(2)Determine the number,magnitude and position of the image charges.3 Boundary Value ProblemsThe number of image charges is3 Boundary Value Problems 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.Compute the surface charge density on the surface of the conducting sphere.3 Boundary Value Problems 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 is3 Boundary Value Problems From the boundary conditions ,we obtain Squaring both sides This is an identity with respect to .Thus,Solving these equations,we get and3 Boundary Value Problems 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 3 Boundary Value ProblemsThus,the total charge induced on the surface is3 Boundary Value Problems图 球外的电场分布3 Boundary Value Problems（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.qdaConsider图 点电荷位于不接地导体球附近的场图3 Boundary Value Problems 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)3 Boundary Value Problems Solution The boundary conditions are and at the plane Region 1:Place an imaginary charge at(0,0,-d).We temporarily assume the permittivity of medium 2 is also The potential at any point P in region1 is 3 Boundary Value Problems 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 3 Boundary Value ProblemsAny point on the interface satisfiesFrom boundary conditions,we haveSolving these equations,we obtain3 Boundary Value ProblemsThus,the potential at any point in region 1 isThe potential at any point in Region 2 is3 Boundary Value Problems3 Boundary Value Problems 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 potential in space.3 Boundary Value Problems3 Boundary Value Problems 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.3 Boundary Value Problems Thus,we have3 Boundary Value ProblemsLet We obtain andThus,the potential at any point P is Trial solution试探解3 Boundary Value Problems The surface charge density on the surface of the wire is The charge induced per unit length on the surface isMethod of electric axis3 Boundary Value Problems Example 3.4.4 Two parallel cylindrical conductors of radius a separated a distance 2d form transmission line,as shown in Figure.Calculate the capacitance per unit length of two cylindrical conductors.yx3 Boundary Value Problems 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 isyx(C=0 zero reference at y axis)3 Boundary Value Problems The capacitance per unit length of two cylindrical conductors is When we obtainyx3 Boundary Value Problems3 Boundary Value ProblemsTwo 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 3 Boundary Value ProblemsA set of eccentric circlesThus,we obtainThe equation of equipotential lines iscenterRadius When the radius is acircle3 Boundary Value Problems 根据 ，得到 Ex 和 Ey 分量Lines of forces of E3 Boundary Value Problems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：3 Boundary Value ProblemsAB3 Boundary Value Problems 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 Boundary Value Problems3.5 The Finite-Difference Method(FDM)3 Boundary Value Problems Numerical methods:finite-difference method(FDM),finite element method(FEM),method of moments(MOM).The finite-difference method(FDM)3 Boundary Value Problems 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。Divide the solution domain into finite discrete points and replace the partial differential equation with a set of difference equations.3 Boundary Value Problems 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 mesh surrounding a point 0.3 Boundary Value Problems Let the potential at point 0 be equal to and at the four surrounding points and The potential at point 1,using the Taylor series expansion,is where h is the mesh size.and the potential at point 3 is3 Boundary Value ProblemsThus,we haveOmitting high order terms,we obtain (2)Similarly,we get(3)Adding(2)and(3),we obtain(4)3 Boundary Value ProblemsSubstituting(1)into(4),we have Thus,the potential at point 0 is In charge-free region,the potential at point 0 is The potential at a point must be the average of the potential at the four surrounding point.3 Boundary Value Problems Example A rectangular trough is formed by four conducting planes.The rectangular trough is infinitely long in the z-direction.The sides and bottom of the trough are at zero potential.The lid is at a potential of 100V.We divide the region into 16 squares.3 Boundary Value Problems3 Boundary Value Problems In iterative method,the potential for the(n+1)th iteration at node(k j,)is 3 Boundary Value Problems Successive over-relaxation(SOR)iterative method where is called acceleration factor.Gauss-Seidel iterative method 高斯赛德尔迭代法逐次超松弛迭代法 迭代过程直到节点电位满足 为止。收敛速度与电位初始值及网格剖分粗细有关；迭代次数与工程精度 有关。3 Boundary Value Problems边界节点赋已知电位值赋节点电位初始值累计迭代次数 n=0n=n+1按超松弛法进行一次迭代，求 打印 NoYes程序框图下 页上 页返 回3 Boundary Value Problems上机作业要求：1.试用超松弛迭代法求解接地金属槽内电位的分布。给定边值：如图示；已知：计算：迭代次数 N=?,分布。给定初值：误差范围：下 页上 页返 回图 接地金属槽的网格剖分3 Boundary Value Problems有限差分法有限元法边界元法矩量法积分方程法积分法分离变量法镜像法、电轴法微分方程法保角变换法计算法实验法解析法数值法实测法模拟法边值问题
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