電磁場與電磁波第9講靜電場的邊值問題

1、Field and Wave Electromagnetic電磁場與電磁波電磁場與電磁波第第9講講2作業(yè)情況作業(yè)情況1班：人班：人合計(jì)：人合計(jì)：人情況情況:3Review, , , , ,eE D P V C W 300(R R )E (V/m), (V)4R R4R RPq-qV- D0E0CE dl SD ds Q2121221212 ( ) 0 or ) or nttnsnnsaEEEEaDDDD（1122eeVVVWD EdvD Edvw dv2121PPWVVE dlq4Main topic 4. Solution of Electrostatic Problems3. Method

2、 of Images 1. Poissons and Laplaces Equations2. Uniqueness of Electrostatic Solutions4. Boundary-Value Problems in Cartesian Coordinates 5 The relationship between the electric potential V and the electric field intensity E is Taking the divergence operation for both sides of the above equation give

3、s In a linear, and isotropic medium, the divergence of the electric field intensity E isV E2VV EE1. Poissons and Laplaces Equations6The differential equation for the electric potential is2V which is called Poissons equation. In a no free charge (source-free) region, and the above equation becomes20V

4、which is called Laplaces equation. 7 Poissons equation states that the Laplacian (the divergence of the gradient) of V equals / for a simple medium, where is the permittivity of the medium (which is a constant) and is the volume charge density of free charges.Operator, 2 , the Laplacian operator, wh

5、ich stands for “the divergence of the gradient of,” or “”.Since both divergence and gradient operations involve first-order spatial derivatives. Poissons equation is a second-order partial differential equation that holds at every point in space where the second-order derivatives exist. Remarks2V 82

6、22222222aaaaaaxyzxyzVVVVVVxyzxyzVVVVxyz 22222211VVVVrrrrrz22222222111VRVRRVRRRVsinsinsinIn Cartesian coordinates:In spherical coordinates:In cylindrical coordinates:9邊值問題研究方法計(jì)算法解析法積分變換法積分變換法分離變量法分離變量法鏡像法（電軸法）鏡像法（電軸法）微分方程法微分方程法保角變換法保角變換法實(shí)驗(yàn)法作圖法實(shí)測法模擬法定性定量數(shù)學(xué)模擬法數(shù)學(xué)模擬法物理模擬法物理模擬法數(shù)值法有限差分法有限差分法有限元法有限元法邊界元法邊界元

7、法矩量法矩量法半解析法/半數(shù)值法格林函數(shù)法格林函數(shù)法10Example 1.一維泊松方程的解一維泊松方程的解The two metal plates having an area A and a separation d form a parallel-plate capacitor. The upper plate is held at potential of V0 , and the lower plate is grounded. Determine(a)the potential distribution(b)the electric field intensity(c)the ch

8、arge distribution on each plate (d)the capacitance of the parallel-plate capacitor 11Solution:1. Choose an appropriate coordinate system for the given geometry2. Governing equation for problems and boundary condition.2222222220VVVVVVxyzz 220Vz區(qū)域中的場方程：000zz dVVV邊界條件勻強(qiáng)電場，電位勻強(qiáng)電場，電位V只是隨高度只是隨高度z的變化而變化的變化

9、而變化-1-0.500.51-1-0.500.5100.511.5zxy1200zzzVVVEVaaDEazdd ；21220VVC zCz 區(qū)域中的場方程：（積分兩次）12122000001010 00 0 zzz dz dVVC zCVVzCCCVVVVVCdC dd4. 特解（帶入邊界條件求解未知系數(shù)）特解（帶入邊界條件求解未知系數(shù)） 3. 方程的通解方程的通解1300;slQVAQAA CdVd 0000The lower plateTh;e up plate;nznnzzslnznnzzsuVVaa DD aaaddVVaa DD aaadd ：0; :nstnDEa導(dǎo)體/介質(zhì)分界面

10、邊界條件：；指向?qū)w外部14Example 2. The inner conductor of radius a of a coaxial cable is held at a potential of V0 while the outer conductor of radius b is groundedDetermine(a)the potential distribution between the conductors (b)the electric field intensity(c)the charge density on the inner conductor (d)the c

11、apacitance of the per unit length151. Choose an appropriate coordinate system for the given geometry2. Governing equation for problems and boundary condition. 22222201101000rrVVVVrrrrrzV rVVrrrrrrrVVV =a=b軸對稱的場，且忽略邊緣效應(yīng)（無限長圓柱體）場方程邊界條件：Solution:-1-0.500.51-1-0.500.5100.10.2xyz161112lnCVVrCrrrVCrC4.4.特

12、解（帶入邊界條件求解未知系數(shù)）特解（帶入邊界條件求解未知系數(shù)） 3. 3. 方程的通解方程的通解012121020000ln;0ln/ln(ln( )/ln( )ln( )ln( )/ln(ln( )/ln(ln(rrVVCaC VCbCaaCVCVbbbrVaabVVrVbabbb =a=b）,)）17001210(ln)ln(ln(ln(rzrrrrrVVVEVaaarrzVVCrCCaaaaabrrrrbaVDEabra）18000000The inner conductor;ln(ln(ln(212ln(ln(122ln(ln(nrinrrararrsrarasaaDD aVVVaab

13、bbrraaaaVVQSabbaaaQCbbVaa：）220; :nstnDEa導(dǎo)體/介質(zhì)分界面邊界條件：；指向?qū)w外部191211220nnDDEE dE dVExample 3 The upper and lower conducting plates of a large parallel-plate capacitor are separated by a distance d and maintained at potentials V0 and 0 respectively. A dielectric slab of dielectric r constant and unifor

14、m thickness 0.8d is placed over the lower plate. E and D ?d0.8d00VyxD2D1E2E120(1)(1) 求解區(qū)域：平行板電容器之間的區(qū)域求解區(qū)域：平行板電容器之間的區(qū)域(2)(2) 分區(qū)：由于填充兩種介質(zhì)，因此場量在分界面上會(huì)分區(qū)：由于填充兩種介質(zhì)，因此場量在分界面上會(huì) 發(fā)生突變，因此，分成兩個(gè)子區(qū)域發(fā)生突變，因此，分成兩個(gè)子區(qū)域(3)(3) 建立坐標(biāo)系：豎直向上為建立坐標(biāo)系：豎直向上為y y軸方向，建立坐標(biāo)系軸方向，建立坐標(biāo)系(4)(4) 場分布分析：在兩種介質(zhì)中都是勻強(qiáng)電場，電位場分布分析：在兩種介質(zhì)中都是勻強(qiáng)電場，電位V

15、V只是隨高度只是隨高度y y的變化而的變化而變化變化 V(yV(y) )，而與，而與x x，z z無關(guān)，無關(guān)，(5)(5) 寫出場方程與邊界條件：待求量是兩個(gè)區(qū)域的電位寫出場方程與邊界條件：待求量是兩個(gè)區(qū)域的電位V V1 1 、V V2 2 ，場方程：泊場方程：泊松方程（有源）松方程（有源）oror拉普拉斯方程（無源）拉普拉斯方程（無源） d0.8d00VyxD2D1E2E121區(qū)域區(qū)域1：區(qū)域區(qū)域2：210V 因?yàn)樵谠搮^(qū)域內(nèi)沒有電荷分布220V 因?yàn)樵谠搮^(qū)域內(nèi)沒有電荷分布22122222222111111222222222222222222200VVVVyxzVVVVVVxyzyVVVVVV

16、xyzy，只是 的函數(shù)，因此0d0.8d00VyxD2D1E2E122 1200120.80.8121212120.80.80.80.80 1 ; 2y34yy dydydydydydydVVVVVVVVVnnyy求解區(qū)域的邊界：分界面上的銜接條件：分界面法線方向?yàn)?方向 寫出通解：寫出通解：112234;VC yCVC yC一維邊值問題一維邊值問題BVP電位的邊界條件，兩個(gè)介質(zhì)的銜接條件：電位的邊界條件，兩個(gè)介質(zhì)的銜接條件：2212220;0VVyy 介 質(zhì) 區(qū) 空 氣 域d0.8d00VyxD2D1E2E1231202000000212112112344323323131313131000

17、2210044100230.8d=0.2d4,555;4yy drrrVCCVCyVCd CCCdVC yCdVC y dCCCVCVCCCVVVVVVVCCddVdVV 由表達(dá)式（）由表達(dá)式（）由表達(dá)式（）由表達(dá)式（）即：12004545;rrrVVyy ddVd0.8d00VyxD2D1E2E124121001200110220200011202255;4455;445544yryyyyrrryyryrrnsrrlowernruyrpnVEVayVVVVaaaaydydVVEaEaddaDDVVDdEEdaaDDD 0005544rryyrrVaad d0.8d00VyxD2D1E2E12

18、52222220VVVxyz In rectangular coordinate system, Laplaces Equation for electric potential is( , , )( ) ( ) ( )V xyzX x Y y Z zLet Substituting it into the above equation, and dividing both sides by X(x)Y(y)Z(z), we have 0dd1dd1dd1222222zZZyYYxXXWhere each term involves only one variable. The only wa

19、y the equation can be satisfied is to have each term equal to a constant. Let these constants be , and we have222 , ,zyxkkk4. Boundary-Value Problems in Cartesian Coordinates 260dd222XkxXx 0dd222YkyYy0dd222ZkzZz The three separation constants are not independent of each other, and they satisfy the f

20、ollowing equation0222zyxkkk The three-dimensional partial differential equation is separated to three ordinary differential equations, and the solutions of the ordinary differential equations are easier to obtain. xkxkxxBAxXjjee)(xkDxkCxXxxcossin)(orwhere A, B, C, D are the constants to be determine

21、d.where kx ，ky ，kz are called the separation constants, and they could be real or imaginary numbers. If kx is an real number, The solution of the equation for the variable x can be written as27xxBAxXee)(xDxCxXcoshsinh)(or The solutions of the equations for the variables y and z have the same forms.

22、The product of these solutions gives the solution of the original partial differential equation. The separation constants could be imaginary numbers. If is an imaginary number, written as , then the equation becomesjxkxk The constants in the solutions are also related to the boundary conditions. It

23、is very important to select the forms of the solutions, which depend on the given boundary conditions.28 Example. Two semi-infinite, grounded conducting planes are parallel to each other with a separation of d. The finite end is closed by a conducting plane held at electric potential V0 , and is iso

24、lated from the semi-infinite grounded conducting plane with a small gap. Find the electric potential in the slot constructed by the three conducting planes. Solution: Select rectangular coordinate system. Since the conducting plane is infinite in the z-direction, the potential in the slot must be in

25、dependent of z, and this is a two-dimensional problem. The Laplaces Equation for the electric potential becomes22220VVxydxyV = 0V = 0V = V0O29( , )( ) ( )V xyX x Y yUsing the method of separation of variables, and let The boundary conditions for the electric potential in the slot can be expressed as

26、 In order to satisfy the boundary conditions and , the solution of Y(y) should be selected as( , )0 V xd ( , 0)0V xykBykAyYyycossin)(0(0, )VyV( , )0Vy( , 0)0V x( , )0 V xd 3, 2, 1, ,ndnky From the boundary condition , we have V = 0 at y = 0 , and the constant B = 0. In order to satisfy , the constan

27、t ky should be( , )0 V xd ( , 0)0V x30ydnAyYsin)(We findSince ，we obtain022yxkkdnkxj The constant kx is an imaginary number, and the solution of X(x) should bexdnxdnDCxXee)(Since V = 0 at x , the constant C = 0, andxdnDxXe)( , )esinnxdnV x yCydThenWhere the constant C = AD .31Since V = V 0 at x = 0

28、, and we have0sinnVCyd The right side of the above equation is variable, since C and n are not fixed. To satisfy the requirement at x = 0, one needs to take the linear combination of the equation as the solution, leading to1( , )esinnxdnnnV x yCyd In order to satisfy the boundary condition x = 0, V

29、= V0 , and we have 01sin, 0nnnVCyydddxyV = 0V = 0V = V0O32 The right side of the above equation is Fourier series. By using the orthogonality between the terms of a Fourier series, the coefficients Cn can be found as041( , )esinnxdnVnV x yyndFinally, we find the electric potential in the slot as 5,

30、3, , 1n0dxyV = 0V = 0V = V0Electric field linesEquipotential surfaces04, if is odd 0 , if is even nVnCnn33uniqueness theorem: means that a solution of Poissons equation (of which Laplaces equation is a special case) that satisfies the given boundary conditions is a unique solution. It does not mean

31、that only one method can be used to obtain the solution of the electrostatic problem. The implication of the uniqueness theorem is that a solution of an electrostatic problem with its boundary conditions is the only possible solution irrespective of the method by which the solution is obtained. A so

32、lution obtained even by intelligent guessing is the only correct solution 2. Uniqueness of Electrostatic Solutions3420Ea4RqR2030Ea()4Ea()4RRqRaRqRRaa點(diǎn)電荷和帶電的球殼、球體在點(diǎn)電荷和帶電的球殼、球體在RaRa的區(qū)域中產(chǎn)生的場是是相等的，的區(qū)域中產(chǎn)生的場是是相等的，稱為這三種源是稱為這三種源是相互等效相互等效的的. .注注：在：在RaR0).s(1)chap2：感應(yīng)電荷很難求：感應(yīng)電荷很難求20,QVx yd z (2)直接解方程直接解方程:36y

33、QdHalfspace problemx點(diǎn)電荷感應(yīng)電荷產(chǎn)生的場，靜態(tài)平衡后，導(dǎo)體表面是等勢面，電力線與其點(diǎn)電荷感應(yīng)電荷產(chǎn)生的場，靜態(tài)平衡后，導(dǎo)體表面是等勢面，電力線與其正交。而這種電力線的分布與以正交。而這種電力線的分布與以xozxoz平面為對稱面，在（平面為對稱面，在（0 0，d d，0 0）處點(diǎn)電荷）處點(diǎn)電荷Q Q，（0 0，d d，0 0）處有）處有Q Q 的一對點(diǎn)電荷在的一對點(diǎn)電荷在x0 x0空間的空間的電力線分布相似。電力線分布相似。P(3)(3)另辟蹊徑：另辟蹊徑： ( (等效原理等效原理) )感應(yīng)感應(yīng)（極化）電荷產(chǎn)生的場，由（極化）電荷產(chǎn)生的場，由假想假想的簡單電荷（像點(diǎn)電荷的簡

34、單電荷（像點(diǎn)電荷 線電荷等）分布產(chǎn)生的場來線電荷等）分布產(chǎn)生的場來等效等效(4)(4)問題：問題：引入像電荷后求得的場，是不是原問題的場？引入像電荷后求得的場，是不是原問題的場？判斷的依據(jù)判斷的依據(jù) （uniquenessuniqueness theoremtheorem）是不是滿足原問題的場方程邊界條件？是不是滿足原問題的場方程邊界條件？37Image ChargeImage method V(x,0, z) = 0yQQ( , , )P x y z根據(jù)場疊加原理，寫出點(diǎn)電荷和像電荷在上根據(jù)場疊加原理，寫出點(diǎn)電荷和像電荷在上半空間任意一點(diǎn)半空間任意一點(diǎn) P P 處產(chǎn)生的場的表達(dá)式處產(chǎn)生的場的

35、表達(dá)式011( , , )4QV x y zRR222222;RxydzRxydz,0 xyzV000RRyyVV 因?yàn)?，BVP BC（判斷的條件）（判斷的條件）20,0,;0;0yxyzQVx yd zVV 等效問題的場就是原問題的場等效問題的場就是原問題的場38Method of Image Essence: The effect of the boundary is replaced by one or several equivalent charges, and the original inhomogeneous region with a boundary becomes an

36、infinite homogeneous space. Basis：The principle of uniqueness. Therefore, these charges should not change the original boundary conditions. These equivalent charges are at the image positions of the original charges, and are called image charges, and this method is called the method of images. Key：T

37、o determine the values and the positions of the image charges. Restriction：These image charges may be determined only for some special boundaries (infinite plane, infinitely long wedge, infinitely long cylindrical, and spherical boundaries) and charges with certain distributions.39 q For the semi-in

38、finite wedge conducting boundary, the method of images is also applicable. However, the images can be found only for conducting wedges with angle given by where n is an integer. In order to keep the wedge boundary at zero-potential, several image charges are required. n/3 When an infinite line charg

39、e is nearby an infinite conducting plane, the method of images can be applied as well, based on the principle of superposition./3q40summary1. Poissons and Laplaces Equations2V 20V222222222aaaaaaxyzxyzVVVVVVxyzxVyzxyz In Cartesian coordinates:413. Method of Images 4. Boundary-Value Problems in Cartesian Coordinates 2. Uniqueness of Electrostatic Solutionsuniqueness theorem: means that a solution of Poissons equation (of which Laplaces equation is a special case) that satisfies the given boundary conditions is a unique solution. Essence: The effect of the boundary i