E-di电势 Electric potential(rs)-Φ(ra)=-静电场强在两点间的线积分与路径无关.Thelineintegrationofanelectricfieldbetweentwopoints is independentofthepath.静电场是保守场.Electricfieldisconservative电势是位置的标量函数.Thepotential isascalarfunctionofposition.Onlydifferencesof potential are defined.For an isolated system of charges it is usual for the potential at infinityto be chosen as zero.q孤立点电荷的电势Potential ofanisolatedpointchargeΦ(r)=24TE0rL点电荷系统的电势Potntialofasystemofpintchages:中(r)=4元r-r,电场强度作为电势梯Electricfieldasthegradientofthepotential(Differentialrelationbetween electric field and potential)[radE=-=-Φ=-1111azaxThe electric field at a certain point is equal to the minus of the gradient of the potential.The electric potential is a scalar function.The gradient of the potential is a vectorfunction.等势面Equipotential surfaceLine of force and equipotential surface are always perpendicular偶极电势 Dipole potential (r)= Pcosp-r4E04TTE0Thedipolepotential andfield are importantbecause theyapply to real atoms电场能量Energyofanelectricfield:010U=CI电容器储藏的能量Energy stored by an capacitorU平板容器的电eaianoarallatcaiordV电场能量用电场强度表达Energyofanelectricfieldintermsoftheelectricfield（storedU=JSE'dTin a volume V)Thisformula can be applied to the electric field produce by any charge system.6
电势 Electric potential r B −r A =−∫A B E⋅d l 静电场强在两点间的线积分与路径无关. The line integration of an electric field between two points is independent of the path. 静电场是保守场. Electric field is conservative. 电势是位置的标量函数. The potential is a scalar function of position. Only differences of potential are defined. For an isolated system of charges it is usual for the potential at infinity to be chosen as zero. 孤立点电荷的电势 Potential of an isolated point charge r= q 40 r 点电荷系统的电势 Potential of a system of point charges: r= 1 40 ∑i qi ∣r−ri ∣ 电场强度作为电势梯 Electric field as the gradient of the potential (Differential relation between electric field and potential) E=−∇ =−∇ =−[ i ∂ ∂ x j ∂ ∂ y k ∂ ∂ z ] The electric field at a certain point is equal to the minus of the gradient of the potential. The electric potential is a scalar function. The gradient of the potential is a vector function. 等势面 Equipotential surface: Line of force and equipotential surface are always perpendicular. 偶极电势 Dipole potential r= pcos 40 r 2 = p⋅r 40 r 3 The dipole potential and field are important because they apply to real atoms. 电场能量 Energy of an electric field: 电容器储藏的能量 Energy stored by an capacitor U = 1 2 CV 2= 1 2 Q 2 /C 平板电容器的电容 The capacitance of a parallel plate capacitor C= Q V = 0 A d 电场能量用电场强度表达 Energy of an electric field in terms of the electric field (stored in a volume V) U=∫V 1 2 0 E 2 d This formula can be applied to the electric field produce by any charge system. 6
电容器极板间的力JTheforcebetweentheplatesof ancapacitordU-1F-oE" A. The force equals the gradient of the energy.dx2习题类型求电场：用点电荷电场的矢量叠加。用高斯定理（有对称性的情形，如无穷大平面，无穷长柱体，球）。用电势梯度。已知击穿电压求最高电势。求电势：用电场强度的积分。用点电荷电势的叠加。求静电力：若干点电荷间的力，用库仑定律：或，力等于电场强度乘以受力电荷的电量。电容器的计算：求电势，电场强度，电荷密度，灵活运用公式。施于电容器极板上的力，用能量关系。第二章：电介质概要1、教学内容：2.1极化2.2相对电容率和极化率，2.3电介质中的宏观场2.4存在电介质情形的静电能2、教学基本要求：掌握极化、极化强度、电位移矢量、介质中的高斯定理、边界条件、介质中的静电能偶极矩在外场中的能量，受力、力矩。3、重点：极化强度、电位移矢量、介质中的高斯定理。难点：极化强度、电位移量的概念及计算、偶极矩在外场中的能量，受力、力矩。7
电 容 器 极 板 间 的 力 The force between the plates of an capacitor: F= dU dx = 1 2 0 E 2 A . The force equals the gradient of the energy. 习题类型 求电场：用点电荷电场的矢量叠加。用高斯定理（有对称性的情形，如无穷大平面，无 穷长柱体，球）。用电势梯度。已知击穿电压求最高电势。 求电势：用电场强度的积分。用点电荷电势的叠加。 求静电力：若干点电荷间的力，用库仑定律；或，力等于电场强度乘以受力电荷的电量。 电容器的计算：求电势，电场强度，电荷密度，灵活运用公式。施于电容器极板上的力， 用能量关系。 第二章：电介质 概要 1、教学内容： 2.1 极化 2.2 相对电容率和极化率 2.3 电介质中的宏观场 2.4 存在电介质情 形的静电能 2、教学基本要求： 掌握极化、极化强度、电位移矢量、介质中的高斯定理、边界条件、介质中的静电能， 偶极矩在外场中的能量，受力、力矩。 3、重点：极化强度、电位移矢量、介质中的高斯定理。 难点：极化强度、电位移矢量的概念及计算、偶极矩在外场中的能量，受力、力矩。 7
基本概念、定律和公式电介质Dielectrics:insulatingmaterials,inwhichtheappearanceofinducedchargesreducethe field in the insulator原子的电偶极矩Dipolemoment ofthe polarizedatom:p=Zea，aisthedisplacementof the electron cloud极化强度矢量Polarization:P=NpThedipole moment per unit volume of the dielectrics.外场中的电偶极矩Dipoleinanexternalfield:电偶极矩在外场中的能量Theenergyofandipoleinexternalfield:U=-qaEcoso=-p·E电偶极矩在外场中受的力Theforceonandipolebyexternalfield:F=-VU=V(P·E)电偶极矩在外场中受的力矩Thetorqueonandipolebyexternalfield:T=P×E相对电容率（相对介电常数）RelativepermittivityEIn dielectrics, the field is reduced by a factorC-EEAE=E=0l0Incapacitors,d电极化率electric susceptibilityXe：X=E-1P=(=-1)E,EThe relation between polarization and electric field is P=XgEE.宏观极化电荷密度Macroscopicpolarizationchargedensity:aPxap,oP:V.PPpax0zOyV.D=PF,D=EEE电位移矢量Electricdisplacementvector:D=E+P8
基本概念、定律和公式 电介质 Dielectrics: insulating materials, in which the appearance of induced charges reduce the field in the insulator. 原子的电偶极矩 Dipole moment of the polarized atom: p=Ze a , a is the displacement of the electron cloud. 极化强度矢量 Polarization: The dipole moment per unit volume of the dielectrics. P=N p 外场中的电偶极矩 Dipole in an external field: 电偶极矩在外场中的能量 The energy of an dipole in external field: U =−qaE cos=− p⋅E 电偶极矩在外场中受的力 The force on an dipole by external field: F=−∇U=∇P⋅E 电偶极矩在外场中受的力矩 The torque on an dipole by external field: =P×E 相对电容率（相对介电常数）Relative permittivity : In dielectrics, the field is reduced by a factor . In capacitors, E=E0 /=/ 0 C= 0 A d . 电极化率 electric susceptibility E ： E=−1 . The relation between polarization and electric field is P=E 0 E , P=−10 E . 宏观极化电荷密度 Macroscopic polarization charge density: P=− ∂Px ∂ x − ∂ Py ∂ y − ∂ Pz ∂ z =−∇⋅P . 电位移矢量 Electric displacement vector: D=0 EP , ∇⋅D=f , D= 0 E . 8
介质中的高斯定理Gauss'lawinthepresenceofdielectricsFlux of D out ofa closed surface equals to total free charge enclosed within the surfaceJ,Dds=J,pfdt电场和电位移的边界条件BoundaryconditionforEandDE//=E2ll，D21=Di1，or D21-D1=0D.Edt存在电介质情形的能量(Energy inthepresence of dielectrics)：U=习题类型求介质中的电极化强度，总电场，电位移，先求D，再求E，P。计算含有介质的电容器：电场强度，电势，电荷密度。偶极矩在外场中的能量，受力，力矩，用相应的公式。第三章：静电场的计算概要1、教学内容：3.1引言3.2具有简单对称性的电场3.3泊松方程3.4不同区域的边界3.5镜像法3.6存在电介质情形的电势分布3.7球坐标系中的分离变量法2、教学基本要求：掌握静泊松方程、唯一性定理、边界条件的应用、电像法、解一维泊松方程、解具有轴对称的球边值问题。3、重点：运用边界条件解一维泊松方程、电像法。9
介质中的高斯定理 Gauss' law in the presence of dielectrics: Flux of D out of a closed surface equals to total free charge enclosed within the surface. ∫S D⋅d S=∫V f d . 电场和电位移的边界条件 Boundary condition for E and D: E1||=E2 || , D2 ⊥= D1 ⊥ , or D2 ⊥− D1 ⊥= f . 存在电介质情形的能量 (Energy in the presence of dielectrics): U = 1 2 ∫V D⋅Ed 习题类型 求介质中的电极化强度，总电场，电位移，先求 D，再求 E，P。 计算含有介质的电容器：电场强度，电势，电荷密度。 偶极矩在外场中的能量，受力，力矩，用相应的公式。 第三章：静电场的计算 概要 1、教学内容： 3.1 引言 3.2 具有简单对称性的电场 3.3 泊松方程 3.4 不同区域的边界 3.5 镜像法 3.6 存在电介质情形的电势分布 3.7 球坐标系中的分离变量法 2、教学基本要求： 掌握静泊松方程、唯一性定理、边界条件的应用、电像法、解一维泊松方程、解具 有轴对称的球边值问题。 3、重点：运用边界条件解一维泊松方程、电像法。 9
难点：运用边界条件解一维泊松方程、电像法、解具有轴对称的球边值问题。基本概念、定律和公式泊松方程Poisson'sequationV=-pleo, where p is total charge density.V'=-P /eE。 where p is free charge density.V2拉普拉斯算符Laplacianoperator=++inCartesian coordinates,ox0y()++in cylindrical polar coordinates,rararl2a020for a long cable with cylindrical symmetry, there are no e and z dependence1 ddΦ(r)=0the Laplace's equation reduces tor dr (dr11a2aasinein spherical coordinatessinede0eorPsin'e ap?Iftheproblemhasazimuthalsymmetry，oraxialsymmetry(方位角对称，或轴对1ad2003sine称),theLaplace'sequationreducestoaoarsingge唯一性定理TheuniquenesstheoremIf Poisson's equation or Laplace's equation has specified values on the boundaries ofsome region, the solution to the equation in this region is the only possible solution with theseboundary values.一维拉普拉斯方程的解SolutionstoonedimensionalLaplace'sequations:1 d(d)b(r)=0 ,solution Φ(r)=Aln(r)+CEquationrdrdr10
难点：运用边界条件解一维泊松方程、电像法、解具有轴对称的球边值问题。 基本概念、定律和公式 泊松方程 Poisson's equation: ∇ 2=−/0 , where is total charge density. ∇ 2=− f / 0 , where is free charge density. 拉普拉斯算符 Laplacian operator ∇ 2 : in Cartesian coordinates, ∇ 2= ∂ 2 ∂ x 2 ∂ 2 ∂ y 2 ∂ 2 ∂ z 2 in cylindrical polar coordinates, ∇ 2= 1 r ∂ ∂ r r ∂ ∂ r 1 r 2 ∂ 2 ∂ 2 ∂ 2 ∂ z 2 for a long cable with cylindrical symmetry, there are no and z dependence, the Laplace's equation reduces to 1 r d dr r d dr r=0 . in spherical coordinates: ∇ 2= 1 r 2 r 2 ∂ ∂r 1 r 2 sin ∂ ∂ sin ∂ ∂ 1 r 2 sin2 ∂ 2 ∂ 2 . If the problem has azimuthal symmetry, or axial symmetry(方位角对称，或轴对 称), the Laplace's equation reduces to r 2 ∂ ∂ r 1 sin ∂ ∂ sin ∂ ∂ =0 . 唯一性定理 The uniqueness theorem: If Poisson's equation or Laplace's equation has specified values on the boundaries of some region, the solution to the equation in this region is the only possible solution with these boundary values. 一维拉普拉斯方程的解 Solutions to one dimensional Laplace's equations: Equation 1 r d dr r d dr r=0 , solution r=AlnrC . 10