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1、3 Distribution of Chargea)All of the charges is on the surface of the conductor Prove：b)The electric field on the surface of a conductor sNote:The field at the surface of the conductor(and just outside)c)Distribution of Charge of Spherical ConductorRestricting factors：the electric field is zero insi
2、de a conductor.第1页/共23页E=0q4Conductor shell（腔）(1)No charge in a conductor shell a The electric field is zero inside inside conductor and shell 导体内、空腔内场强皆为零；b The conductor and shell are equipotential导体、空腔为等势体；c The charge must lie on the surface 内表面处处没有电荷，电荷只分布在外表面.(2)A charge q in a conductor shell
3、 Ontheinnersurfaceoftheshellthechargeis-q.On the outer surface of the shell the charge is q.It is determined by Gausss law.第2页/共23页5 Electrostatic Screening 静电屏蔽Electrostatic Screening:Insulating the influence of electric field using conductor shell No matter the change of charge A the force on B ch
4、arge always is zero.A电荷变化,B电荷受力总是为零。The charge inside shell can be shielded 壳可以屏蔽内部If the charge B is changed,the force on the charge A is changed too.B电荷变化，A电荷受力会变化。The charge A outside shell can not be shielded 壳不能屏蔽外边电荷A。A。BIf the shell is the ground,the influence between A and B is not exist.接地导
5、体壳，可以使壳内外的电荷彼此不影响。第3页/共23页例：球形壳C，电荷A带q,B带Q，r 远大于R。求:A，B，C彼此之作用力。解：(1)A、B之间作用力：(2)A、C对B来说是带电量q的点电荷，B受力：(3)静电屏蔽，A受力 F=0+QBA+qCrR(4)A、C之间作用力：(5)B、C之间作用力F0(6)C接地，A、B受力为零问题：导体壳是否真的能够屏蔽掉外电场使其不能进入腔内？答：接地导体壳return第4页/共23页电介质中不存在自由电荷，在电场中会发生极化现象。极性介质：HCl、SO2、H2O 非极性介质：O O2 2 H H2 2 CO CO2 21 Dielectric polar
6、ization无外电场时，电偶极矩取向杂乱或为零，整个介质呈电中性。当有外电场时，非极性分子产生极化，极性分子趋向外场排列，介质表面产生束缚电荷。E-2 Polarization 极化的强弱程度用电极化强度矢量描述。由于每个分子为偶极子，偶极矩P越大，单位体积内偶极子越多，极化越强。线性各向同性介质2 Dielectrics and Dielectric Polarization 第5页/共23页3 The relationship between electric polarization and bound charge4 Electric displacement Gausss law
7、in mediumElectric displacement 电位移矢量Relative permittivity 相对电容率The field in medium 介质中的场：(Absolute)permittivity 电容率第6页/共23页解：QR例1一点电荷Q被半径为R的介质1包围，球外是2的介质，求空间电场强度。例2平行极板间极化强度矢量，求介质表面束缚电荷密度解：选取圆柱形高斯面return第7页/共23页110-310-610-910-12FmFF微nF纳pF皮1 The capacitance of an Isolated ConductorUnit：Farads法拉Thera
8、diusoftheearthR6370kmC=7.0810-4F3 Capacitance Capacitor Capacitance is a measure of amount of charge a conductor holds at a given potential difference 2 Capacitance of a CapacitorCapacitor：Itisconsistsoftwoconductorplatesbeingveryclosed第8页/共23页3 Combinations of Capacitors 电容器的联接(1)Connected in Serie
9、s 串联(2)Connected in Parallel 并联Forcapacitorsconnectedinseriesthechargeoneachcapacitoristhesame.C1C2C1C2CapacitorsconnectedinParallelhavethesamepotentialdifferenceacrosstheirplateComparing 弹簧的串并联，电阻的串并联，物理中由几个相同量纲量组成另一个同量纲量时，只有上面的组合。第9页/共23页4 Calculating the Capacitance of a Capacitor(1)A Parallel-pl
10、ate CapacitorThecapacitanceofacapacitordependsontheseparationdistanced,theareaSandthepermittivityRARBq-qSD=0-QdABQ(2)A Spherical Capacitor第10页/共23页(3)A Cylindrical Capacitor-取高为h 圆柱形高斯面RARBr hNote：1 Thecapacitanceofacapacitordependsonthegeometryofthecapacitorandthepermittivity,independentof thepoten
11、tialdifferenceandthecharge2 Thesphericalcapacitorisastandardcapacitor第11页/共23页Ex1 A parallel-plate capacitor consists of two square plates,each of sides 4.0 cm separated by 2.0 mm.(a)Calculate the capacitance.(b)How many electrons can be stored on a plate if the capacitor is charged to 50 V?Solution
12、(a)(b)Q=CU=7.110-1250=3.510-10CThemagnitudeofthechargeofoneelectronis1.610-19C,sothenumberNofelectronsthatcanbestoredis=7.110-12F=7.1pF.第12页/共23页Ex2Asphericalcapacitorconsistsoftwothinconcentricsphericalshellsofradiiaandb,withab.Acharge Qisplacedontheinnershell,andQ ontheoutershell,whichisgrounded.C
13、alculatethecapacitance.InordertouseC=,weneedfirsttocalculatethepotentialdifferencebetweentheshells,whichinturndependsonthefield.Aconcentricsphereofradiusr,wherearEB DA=DBreturn第16页/共23页5 Application例1无限长均匀带电圆柱面内外的场强分布：0，R面内面外方向垂直柱面向外解：电荷分布具有柱对称性，电场分布具有柱对称性；利用高斯定理求场强分布，选取柱形高斯面rR第17页/共23页例2无限长均匀带电圆柱体内
14、外的场强分布：0，R柱体内柱体外方向垂直圆柱体轴线向外方向垂直圆柱体轴线向外rR第18页/共23页例3无限长均匀带电圆柱面，半径为R，单位长带电量为+，试求其电势分布。解：电场强度分布：rR：rQB：（1）证明两板内侧电荷密度符号相反，数值相等；外侧的电荷密度符号相同，数值相等。（2）计算板上的电荷面密度和两板间的电势差。解：(1)证明（2）电荷守恒定律：ABQAQBdP1P2P1：静电平衡导体内部场强为零A、B间总场：第21页/共23页例6把一块原来不带电的大金属板B移近一块带电Q的A板，B接地和不接地，两极板的电势差是否一样？B接地B不接地ABreturn第22页/共23页感谢您的观赏第23页/共23页