电磁场与电磁波第8讲边界条件电容能量.pptx

11.Electric Potential Review22.Conductors in Static Electric Field3.Dielectrics in Static Electric Field Inside a conductor(under static conditions)Boundary Conditions(at a Conductor/Free Space Interface)34.Electric Flux Density and Dielectric Constant4Main topic1.静电场的边界条件静电场的边界条件 2.电容和电容器电容和电容器3.静电场的能量静电场的能量51.Boundary Conditions for Electrostatic Fields 电电磁磁问问题题通通常常涉涉及及具具有有不不同同物物理理特特性性的的媒媒质质,并并且且需需要要关关于于两两种种媒媒质质之之间间分分界界面面的的场场量量关关系系的的知知识识.例例如如,我我们们希希望望知知道道场场矢矢量量 E 和和 D 在在穿穿过过两两个个界界面面时时是是如如何何变变化化的的.现现在在我我们们来来考考虑虑两种一般媒质之间的分界面两种一般媒质之间的分界面.6(1)电场强度电场强度E的切向分量的切向分量(tangential component)Let us construct a small path abcda with sides ab and cd in media 1 and 2 respectively,both being parallel to the interface 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 hacdban27 电电场场强强度度在在分分界界面面处处的的切切向向分分量量是是相相等等的的.换换句句话话说说,电电场场强强度度的切向分量连续的切向分量连续.对于线性各向同性的一般介质对于线性各向同性的一般介质,有有:an2 的方向是有媒质的方向是有媒质 指向媒质指向媒质.8(2)电通密度电通密度D的法向分量的法向分量(normal component)In 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 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 s9边界的法线方向由媒质边界的法线方向由媒质 指向媒质指向媒质 .S 是边界上的面电荷密度是边界上的面电荷密度.D场场的的法法向向分分量量在在通通过过存存在在面面电电荷荷的的分分界界面面时时是是不不连连续续的的,不不连连续续的的量量等等于面电荷密度于面电荷密度.10讨论讨论(1)、Boundary Conditions for Dielectric-conductor Interface 讨论讨论(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:11例例1:两两理理想想介介质质的的分分界界面面为为Z=0的的平平面面，如如图所示，在介质图所示，在介质2中的场强为中的场强为求介质求介质1中分解面上的场分量。中分解面上的场分量。2=0 r21=0 r1xzy12132.电容和电容器电容和电容器导体导体比比例例常常数数 C 称称为为孤孤立立导导体体的的电电容容.电电容容是是指指电电位位每每增增加加一一单单位位所所必必须须施施加加于物体的电荷量于物体的电荷量.14 电电容容器器的的电电容容是是双双导导体体系系统统的的一一种种物物理理性性质质,其其依依赖赖于于导导体体的的形形状状和和它它们们之之间间媒媒质质的的介介电电常常数数,与与电电荷荷Q 和和导导体体之之间间的的电电位位差差V12都都无无关关.电电容容器器即即使使没没有有电电压压提提供供或或导导体体上上没没有有自自由由电电荷荷它它也也是是存存在的在的.15Capacitance C can be determined from above equation by either(1)assuming a V12 and determining Q in terms of V12,or(2)assuming 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.对已知的几何形状对已知的几何形状,选择合适的坐标系选择合适的坐标系.2.假设导体上携带了电荷假设导体上携带了电荷+Q and-Q.3.通过高斯定理或其它关系根据假设的电荷量通过高斯定理或其它关系根据假设的电荷量Q来确定来确定 E.4.求出导体两端的电位差求出导体两端的电位差 V12.5.Find C by taking the ratio Q/V12.16Example 3-18 P12417Example 3-19 P12518(1)SeriesParallel Connections of CapacitorsParallel Connections of Capacitors 电容器的电压相等电容器的电压相等 Series connections of capacitors(电容器的电量相等电容器的电量相等)(2)Capacitances In Multiconductor Systems19静静电电能能来来源源：外外力力克克服服电电场场力力做做功功转转化化而而来来，静静电电场场能能仅仅与与带带电电体体的的最最终终带带电电状状态态有有关关而而与与到达这一状态的中间过程无关。到达这一状态的中间过程无关。静静电电能能：当当电电荷荷放放入入电电场场中中，就就会会受受到到电电场场力力的的作作用用，电电场场力力做做功功使使电电荷荷位位移移，这这说说明明电电场场具具有有能能量量。静静电电场场内内储储存存着着能能量量，这这种种能能量量通通常常被被称称为为静静电电能能。电电场场越越强强，对对电电荷荷的的力力就就越越大大，做做功功的的能能力力就就越越强强，说说明明电场具有的能量就越大。电场具有的能量就越大。3.Electrostatic Energy能能量量的的零零点点：最最初初电电荷荷都都分分散散在在彼彼此此相相距距很很远远(无无限限远远)的的位位置置上上。通通常常规规定定，处处于于这这种种状状态态下下的的静静电电能能为为零零。静静电电场场能能量量W We e等等于于于于把把各各部部分分电电荷荷从从无无限限分分散散的的状状态态聚聚集集成成现现有有带带电电体系时抵抗静电力所作的全部功。体系时抵抗静电力所作的全部功。20Bring a charge Q2 from infinity against the field of a charge Q1 in free space to a distance R12(1)Two chargeswhere 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 Q2.21Bring another charge 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(2)Three charges22We can rewrite W3 in the following form where V1 is the potential at Q1 established by charges Q2 and Q3,similarly,V2 and V3 are the potentials at Q2 and Q3,respectively,in the three-charge assembly.23Extending 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 subscript e on We is to denote that the energy is of an electric nature.)We havewhere 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 rest24Two remarks are in order 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).(4)a continuous charge distribution We 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 charge element with all other infinitesimal charge elements.25Example 3-23 P136Find the energy required to assemble a uniform sphere of charge of radius b and volume charge density .RbdR26(4)Electrostatic Energy in Terms at Field Quantities Recalling the vector identity,We can write as27Since V can be any volume that includes all the charges,we may choose it to be a very large sphere with radius R(R).For a linear medium,we have D=E,28We can always define an electrostatic energy density we mathematically,However,this definition of energy density is artificial because a physical justification has not been found to localize energy with an electric field;all we know is that the volume integrals in Equation give the correct total electrostatic energy.29Example 3-24 P138yx电容器储存的总能量：电容器储存的总能量：30summary1.Boundary Conditions for Electrostatic Fields 2.Capacitance and Capacitors313.Electrostatic Energy32homeworkThank you!Bye-bye!Thank you!Bye-bye!答疑安排答疑安排时间：时间：地点：地点：1401，1403P.3-25,3-4433 In addition,we can find the relationship between the bound charges and the normal component of the electric field intensity ashS 2 1an2D1D2 s342)Capacitances In Multiconductor Systems32N1