電磁場與電磁波第8講邊界條件電容能量

1、1Field and Wave Electromagnetic電磁場與電磁波電磁場與電磁波2009. 3. 262作業(yè)情況作業(yè)情況1班：人班：人2班：人班：人合計：人合計：人情況情況:31. Electric Potential EV2121PPWVVE dlq 0 ( )PpV PV PVVE dlV0 (V)4qVR 101 V4RRnkkkqV01 (V)4VVdvR01 (V)4sSVdsR01 (V)4lLVdlRReview42. Conductors in Static Electric Field3. Dielectrics in Static Electric Field I

2、nside a conductor(under static conditions)00E Boundary Conditions(at a Conductor/Free Space Interface)00tsnEE210lim (C/m ) n vkkvpPv psnP apP 54. Electric Flux Density and Dielectric Constant2000(1) (C/m )erDEPEEE = D0E0CE dl SD ds Q6Main topic1. Boundary Conditions for Electrostatic Fields 2. Capac

3、itance and Capacitors3. Electrostatic Energy71. Boundary Conditions for Electrostatic Fields Electromagnetic problems often involve media with different physical properties and require the knowledge of the relations of the field quantities at an interface between two media. For instance, we may wish

4、 to determine how the E and D vectors change in crossing an interface.we now consider an interface between two general media D0E0CE dl SD ds Q8(1) the tangential component of ELet us construct a small path abcda with sides ab and cd in media 1 and 2 respectively, both being parallel to the interface

5、 and equal to w. The integral form is assumed to be valid for regions containing discontinuous media, is applied to this path. If we let sides bc=da= h approach zero, their contributions to the line integral of E around the path can be neglected. Thus we have E1E2 2 1at w hacdban2 121t2t1t2t d d d d

6、 d0 d d()0bcdalabcdbdacEwEwEE 1212ElElElElElElElEwEw9 The tangential components of the electric field intensities in both sides of the interface between two dielectrics are equal. Or say, the tangential components of the electric field intensities are continuous.For linear isotropic dielectrics, we

7、have 22t1t 1DD12ttEE212 ( ) 0 naEEWhere the reference unit normal an2 is ourward from medium to . 10(2) the normal component of DIn order to find a relation between the normal components of the fields at a boundary, we construct a small pillbox with its top face in medium 1 and bottom face in medium

8、 2. The faces have an area S, and the height of the pillbox h is vanishingly small. Applying Gausss law to the pillbox, we havehS 2 1an2D1D2 s21 1212222121n2nd()(sSbottomsidetopnntopbottomnsDdSD dSD dSSD dSDdSD aSDaSaDDSDDSS DS112121n2n() or nssaDDDDThe direction of the normal to the boundary is spe

9、cified as that from dielectric to . where S is the surface density of the free charge at the boundary. The above equation states that the normal component of D field is discontinuous across an interface where a surface charge exists-the amount of discontinuity being equal to the surface charge densi

10、ty. 122121221212 ( ) 0 or ) or nttnsnnsaEEEEaDDDD（討論討論(1)、Boundary Conditions for Dielectric-conductor Interface 討論討論(2)、 In the absence of net surface free charge, one has The boundary conditions that must be satisfied for static electric fields are as follows:211211 0 or or ntnsnsaEEaDD02121221212

11、()0 or )0 or nttnnnaEEEEaDDDD（13例例1:兩理想介質的分界面為兩理想介質的分界面為Z=0的平面，如的平面，如圖所示，在介質圖所示，在介質2中的場強為中的場強為2 (5)xyzEa ya xaz求介質求介質1中分解面上的場分量。中分解面上的場分量。2= 0 r21= 0 r1xzy122()0naEE122()nsaDD112211122222201 (5)(5)(5)()0(|(5)xxyyzzxxyyzzxyzxyzzxyzxxyyzzzxyzxzxEa Ea Ea EDaEaEaEEa ya xazDayaxazaa ya xaza Ea Ea Eaayaxa

12、zaEa011|)0yyzzzEaE141212121212121()()(5)0()()(5)0()()0(5)05;5zxxyyzzzxxyyzzyxxyzxyzxyzaayEaxEaEaayEaxEaEayEaxEEEy Ex EEa ya xa 15 Example. A conducting sphere of radius r1 and with positive charge q is enclosed by a conducting spherical shell of internal radius r2. The permittivity of the dielectric

13、 between the sphere and the shell is 1, and the external radius of the shell is r3 . The spherical shell is covered by a dielectric of 2 with external radius r4. The outer region is vacuum. r1r2r3r4 0 2 1 Solution: In view of the spherical symmetry of the structure and the fields, Gauss law can been

14、 applied. Find: (a) The electric field intensities in every region. (b) The free charges on each surface. Taking the spherical surfaces as Gaussian surfaces, it can be seen that the electric field intensities are perpendicular to them.16 In the regions r r1 and r2r r3 , E = 0, since electrostatic fi

15、eld cannot exist in conductors. r1r2r3r4 0 2 11214rqarE2224rqarEFor the same reason, in the region r3r r4 ,In the region r1r r2 , due to dSqDS we have17 Based on the B-C，we can find the free charges on each surface as follows:r = r1:214 rqSr = r4:0Sr = r2:2224 rqSr = r3:2334 rqSr1r2r3r4 0 2 1212 ) n

16、saDD（182. Capacitance and Capacitors導體導體QVEV01 (V)4sSVdsR SSQkQQkQkVkVVkV不變 SSQkQVkVEkEkQkQVkV不變0snE (c/v or F)QCVWhere the constant of proportionality C is called the capacitance of the isolated conducting body. The capacitance is the electric charge that must be added to the body per unit increase

17、 in its electric potential.1912 (c/v or F)QCVV12 mean the potential difference between the two conductors. The capacitance of a capacitor is a physical property of the two-conductor system. It depends on the geometry of the conductors and on the permittivity of the medium between them; it does not d

18、epend on either the charge Q or the potential difference V12. A capacitor has a capacitance even when no voltage is applied to it and no free charges exist on its conductors.20Capacitance C can be determined from above equation by either (1) assuming a V12 and determining Q in terms of V12, or (2) a

19、ssuming a Q and determining V12 in terms of Q. At this stage, since we have not yet studied the methods for solving boundary-value problems (which will be taken up in Chapter 4), we find C by the second method. The procedure is as follows:1. Choose an appropriate coordinate system for the given geom

20、etry. 2. Assume charges +Q and -Q on the conductors. 3. Find E from Q by Gausss law, or other relations. 4. Find V12 by evaluating from the conductor carrying -Q to the other carrying +Q. 5. Find C by taking the ratio Q/V12.1122VE dlsnE21Example 3-18 P12422Example 3-19 P12523(1) SeriesParallel Conne

21、ctions of CapacitorsParallel Connections of Capacitors 電容器的電壓相等電容器的電壓相等 nCCCC21/Series connections of capacitors(電容器的電量相等電容器的電量相等)nnsrCCCC11111(2) Capacitances In Multiconductor Systems24QqP靜電能來源：靜電能來源：外力克服電場力做功轉化而外力克服電場力做功轉化而來來，靜電場能僅與帶電體的最終帶電狀態(tài)有關而與靜電場能僅與帶電體的最終帶電狀態(tài)有關而與到達這一狀態(tài)的中間過程無關。到達這一狀態(tài)的中間過程無關。靜電能

22、：當電荷放入電場中，就會受到靜電能：當電荷放入電場中，就會受到電場力電場力的作用，電場力做功使電的作用，電場力做功使電荷位移，這說明電場具有能量。靜電場內儲存著能量，這種能量通常被荷位移，這說明電場具有能量。靜電場內儲存著能量，這種能量通常被稱為靜電能。電場越強，對電荷的力就越大，做功的能力就越強，說明稱為靜電能。電場越強，對電荷的力就越大，做功的能力就越強，說明電場具有的能量就越大。電場具有的能量就越大。 212121= () PPWWQE dlQ VV 3. Electrostatic Energy能量的零點能量的零點： 最初最初電荷電荷都分散在彼此相距很遠都分散在彼此相距很遠( (無限遠

23、無限遠) )的位的位置上。通常規(guī)定，處于這種狀態(tài)下的靜電能為零置上。通常規(guī)定，處于這種狀態(tài)下的靜電能為零。靜電場能。靜電場能量量W We e等于于把各部分電荷從無限分散的狀態(tài)聚集成現有帶電等于于把各部分電荷從無限分散的狀態(tài)聚集成現有帶電體系時抵抗靜電力所作的全部功。體系時抵抗靜電力所作的全部功。25Bring a charge Q2 from infinity against the field of a charge Q1 in free space to a distance R12120122224RQQVQW111202124VQRQQW)(2211221VQVQW(1) Two ch

24、arges1Q2Q12R1P2Pwhere V2 is the potential at P2 established by charge Q1, chose the reference point for the potential at infinity; This work is stored in the assembly of the two charges as potential energy. Another form where V1 is the potential at P1 established by charge Q2 .26Bring another charge

25、 Q3 from infinity to a point that is R13 to charge Q1 and R23 from charge Q2 in free space ,an additional work is required that equals where V3 is the potential at P3 established by charges Q1 and Q2 , W3 , which is stored in the assembly of the three charges Q1 , Q2 , and Q3 , is 2302130133344RQRQQ

26、VQW23321331122102341RQQRQQRQQWWW(2) Three charges1Q2Q12R1P2P3Q3P13R23R27We can rewrite W3 in the following form )(33221123021301323031201213031202132144444421VQVQVQRQRQQRQRQQRQRQQW23321331122102341RQQRQQRQQWWWwhere V1 is the potential at Q1 established by charges Q2 and Q3 , similarly, V2 and V3 are

27、 the potentials at Q2 and Q3 , respectively, in the three-charge assembly. 31 122331()2WQVQ VQV28Extending this procedure of bringing in additional charges, we arrive at the following general expression for the potential energy of a group of N discrete point charges at rest. (The purpose of the subs

28、cript e on We is to denote that the energy is of an electric nature.) We have112NekkkWQ V1014Njkjjkj kQVRwhere Vk , the electric potential at Qk, is caused by all the other charges and has the following expression:(3) A group of N discrete point charges at rest29112NekkkWQ VTwo remarks are in order

29、here. First, We can be negative. In that case, work is done by the field (not against the field). Second, We in this equation represents only the interaction energy (mutal energy) and does not include the work required to assemble the individual point charges themselves (self-energy). 1014Njkjjkj kQ

30、VR(4) a continuous charge distribution 12eVWVdvWe replace Qk by dv and the summation by an integration and obtain:Note that We in this equation includes the work (self-energy) required to assemble the distribution of macroscopic charges, because it is the energy of interaction of every infinitesimal

31、 charge element with all other infinitesimal charge elements.30Example 3-23 P136Find the energy required to assemble a uniform sphere of charge of radius b and volume charge density . RbdR12eVWVdv31(4) Electrostatic Energy in Terms at Field Quantities 11()22eVVWVdvD VdvRecalling the vector identity,

32、(),VDVDDV111()()2221122eVVVnSVWVDDV dvVD dvDVdvVD a dsD EdvWe can write as321122enSVWVD a dsD EdvSince V can be any volume that includes all the charges, we may choose it to be a very large sphere with radius R (R ).2211,when the first integral will vanishVDSRRRR，.1 (J)2eVWD EdvFor a linear medium, we have D= E,21 (J)2eVWE dv21 (J)2eVDWdv331122eeVVVWD EdvD Edvw dvWe can always define an electrostatic energy density we mathematically, 22311 (J/cm )222eDwD EEHowever, this definition of energy density is artificial because a physic