大學(xué)物理雙語版奧本漢姆課件chap19to21efie1d1

1、Chapter 19 To 21 Electrostatic Field in Vacuum Electric Field and Its Principle of Superposition (疊加疊加) Gausss Law and Its Applications Electric Potential and Its Principle of Superposition Loop-law and Its Applications With the torsion balance (扭稱實(shí)驗(yàn)扭稱實(shí)驗(yàn)) )Coulomb (French scientist) in 1785 was able

2、 to establish the Coulombs law of electrostatic interaction: The force between two point charges at rest was inversely proportional to the square of the distance between them and proportional to the product of the charges. 12 2 12 21 1221 r r qq kFF x 0 y z r1 r2 r12 +q1 +q2 F21 F12 19-1. Coulombs L

3、aw (P460) k is a proportionality constant. The choice of the numerical value and dimensions of k is arbitrary. 229 0 /CmN1099.8 4 1 k N)/(mC1085. 8 2212 0 r r qq F | 4 1 2 21 0 Dielectric constant in vacuum ( (真空介電常量真空介電常量 真空電容率真空電容率) ) 單位制的單位制的 有理化有理化 r d q1 q2 A “point charge” should be: rd (1) It

4、 is an exp. Law, holds both for macroscopic here r is the distance between the point in question and the dipole center (P477). Explanations: dqp electric dipole moment + q + q p d (b) The product qd, involves two intrinsic properties q , inward the closed surface. 0d A 0d A Karl Friedrich Gauss (177

5、71835), German mathematician and physicist. He made a lot of contributions in the fields of experimental physics, theoretical physics and mathematics. He made major contributions to the theory of electromagnetism. 3. Gauss law (高斯定理高斯定理) (P489) 1 d A AE q Er r 2 0 4 To find out the relation between

6、the electric field and its source, consider a very simple case of a positive point charge q at the center of a spherical shell: A q rA r 2 2 0 d 4 E E A q A r 2 2 0 d 4 0 2 2 0 4 4 q r r q The same result for any shaped-closed surface A2 due to the continuity of electric field line. +q If q is outsi

7、de of the area, inward=outward, =0 0 interior surface closed d q AE Gausslaw for vacuum. The relation of the source of electric filed and the field 反映反映場場和和源源 的關(guān)系的關(guān)系. The flux of the electric field through a closed surface of any shape equals 1/ 0 times of the algebraic sum of charges enclosed withi

8、n the surface. 真空中靜電場內(nèi)通過真空中靜電場內(nèi)通過任意閉合曲面任意閉合曲面的電通量等于該的電通量等于該 曲面所包圍的電量的代數(shù)和的曲面所包圍的電量的代數(shù)和的1/ 0倍倍. . (a) Where is the algebraic sum of all interior charges enclosed in the Gaussian surface. is not related to the way of distribution and outside charges. interior q (b) The quantity on the left side of abov

9、e equation is the electric field resulting from all charges, both those inside and those outside the Gaussian surface. E Descriptions: 0 interior surface closed d q AE 4. Applying Gauss lawcalculating (P491)E (d) It gives a simple way to calculate the distribution of electric field for a given charg

10、e distribution with sufficient symmetry. (c) Gausss Law and Coulombs law are equivalent (see P496). However, Gausss Law is hold for the produced by moving charges; Coulombs law is only true for electrostatic field. E How do we choose the surface for calculating the electric flux or ?E A uniformly ch

11、arged spherical volume （帶電球（帶電球 體）體）, radius R and total charge Q. Find the at any point of inside and outside region. E Example 20-4 (P492): R ? Question R Solution: E R r To calculate E outside of the sphere, we choose for Gauss surface (G. S.) as an imaginary spherical shell A1 with r R: )( 4 2 0

12、 Rrr r Q E 1 d A AE using Gausslaw S AE d QrE 0 2 1 4 To calculate E inside of the sphere, we choose for G.S. as an imaginary spherical shell with A2 ,r r). E A portion of a thin, infinite, non- conducting sheet (無限大均勻帶電平面無限大均勻帶電平面) with a uniform (positive) surface charge density . Find the a dista

13、nce r in front of the sheet. E EAAEAEAE ASA 2ddd rightleft 0 2 A EA 0 2 E(sheet of charge) Planar symmetry 垂直板垂直板 面向外面向外, 距板同遠(yuǎn)處距板同遠(yuǎn)處E大小相同大小相同 Solution: Example 20-6 (P493): E E 00 2 2 EEEC 0,0 BA EE ABC ABC Discussion: If we arrange two infinite plates, with uniform opposite , , to be close to each

14、other and parallel as the figure shown. What are the in the region of A, B and C? E The electric field between parallel- plate capacitor in circuit (直流電路中直流電路中 的平行板電容器間的場強(qiáng)的平行板電容器間的場強(qiáng)) is such a case. The symmetry of charge distribution is as same as the one of electric field distribution. Summary: F

15、or certain symmetry arrangements of charge (as illustrated cylindrical, planar and Spherical symmetry), Gausss law is very much easier to use than integration of field components. 1. Features of electrostatic force Coulombs Law for electrostatic force: 2 21 0 4 1 r qq F Newtons Law for gravitational

16、 force 2 21 r mm GF Mathematically identical! Many properties for the gravitational may also be true for electrostatic forces. The work done by electrostatic force is path independent. The electrostatic force is a conservative force. 19-4. Electric Potential (P502) LLL lEqlEqlF0ddd e Based on the ab

17、ove two points, we can infer that the work done on a moving charge around a closed loop in an electrostatic field is zero. The line integral of the electrostatic field around a closed loop is zero (靜電場中場強(qiáng)沿任意靜電場中場強(qiáng)沿任意 閉合環(huán)路的線積分恒等于零閉合環(huán)路的線積分恒等于零) ) 靜電場的靜電場的環(huán)路定理環(huán)路定理. . L lE0d , 0qCirculation theorem ba b

18、 a ab UUlFW d e q E a b Because electrostatic force is a conservative force, an electric potential energy U can be introduced: babaab UUUW 2. Electric potential energy (P515) To ensure U at a point, a reference potential energy should be chosen: 0 UUb then a aa lEqWU d The U of (charge q at arbitrar

19、y point a) = the work done by the during the move of charge q from a to infinity. (q在電在電 場中場中a點(diǎn)的電勢能點(diǎn)的電勢能= =q從從a移到無窮遠(yuǎn)電場力所做功移到無窮遠(yuǎn)電場力所做功).). e F The physical quantity to represent property of field at any point electric potential. 3. Electric potential (P502) aa a a qEl U VEl qq d d (electric potential

20、 at point a) Va equals (in magnitude) work done by the on unit positive charge during the move from a to infinity along any path. (電荷電荷q在電場中某點(diǎn)電勢在電場中某點(diǎn)電勢Va在數(shù)在數(shù) 值上等于值上等于單位正電荷從單位正電荷從a移到無窮遠(yuǎn)處靜電場力所做的功移到無窮遠(yuǎn)處靜電場力所做的功亦等于亦等于 放在放在點(diǎn)點(diǎn)a的單位正電荷所具有的電勢能的單位正電荷所具有的電勢能).). e F When q =1, aaa WUV V is a scalar. It is mea

21、sured in joules per coulomb, or volts (V) (1V=1J/C); dimension is . U is measured in joules (eV in P515). 321 MTLI V is a property of an electric field, regardless of whether a charged object has been placed in that field; U is an energy associated with a system (the charged object plus the charged

22、particles that set up the electric field). (a) The difference of V and U: (b) Unit q0, produces a negative electric potential. (b) Potential Due to a Group of Point Charges n i i r q V 1 0 4 Based on the principle of superposition of :E (d) Principle of superposition of electric potential: (c) Poten

23、tial due to a continuous charge distribution 場場源源電電荷荷 r q VV d 4 1 d 0 V S l q d d d dwhere Assume some charges produces at point a: , 21 EE 2121 dddVVlElElEV aaa 電場中某點(diǎn)的電勢等于各電荷電場中某點(diǎn)的電勢等于各電荷單獨(dú)單獨(dú)在該點(diǎn)產(chǎn)生的電勢的疊加在該點(diǎn)產(chǎn)生的電勢的疊加(代數(shù)和代數(shù)和)。 Note: It should be relative to same reference point. A thin ring of radius

24、R with q uniformly distributing its circumference. What is the V at point P, a distance z from the ring along its central axis (帶電圓環(huán)軸線帶電圓環(huán)軸線) )? l R q lqd 2 dd 2 1 22 0 )(4 )( Rz q zV R L P RzR lq VV 2 0 2 1 22 0 2 )(8 d d Z R r P qd Solution: 2 1 22 0 2 0)(8 d 4 d d RzR lq r q V Example 21-8 (P510)

25、: z V O 球面內(nèi)球面內(nèi)E=0! Calculating potential V for the uniformly charged spherical shell (with charge Q and radius R) . r r Q rV r d 4 )( 2 0 Vin equals to the V at spherical surface. The distribution of V is continuous. Spherical shell is a equipotential body (等勢體等勢體). 與電量集中在球心的與電量集中在球心的 點(diǎn)電荷點(diǎn)電荷的電勢分布相同的

26、電勢分布相同 r r Q rErV R rR d 4 d)( 2 0 )( 4 0 Rr R Q Rr r Q 0 4 Example: V r O R R r 19-5. Relation between V and E 1. Equipotential surfaces (P511) CzyxV ),( Adjacent points that have the same electric potential form an equipotential surface, . 電勢相等的點(diǎn)組成的面電勢相等的點(diǎn)組成的面 Take as a constant and plot equipoten

27、tial surface. V (b) The direction of electric field lines direct the one of electric potential decreasing (電勢降落的方向電勢降落的方向) ) 放在電場中的放在電場中的正電荷將由高電勢處向低電勢處運(yùn)動(dòng)正電荷將由高電勢處向低電勢處運(yùn)動(dòng); ;負(fù)電荷負(fù)電荷 2. Relation between electric field line and equipotential surface (a) Electric field lines are perpendicular to equipoten

28、tial surfaces everywhere. (c) The electrostatic force does zero work if charge move along the surface. (d) The place where the equipotential surfaces are dense has strong electric field (等勢面的等勢面的疏密疏密反映了場的反映了場的強(qiáng)弱強(qiáng)弱). 3. Electric potential gradient (電勢梯度電勢梯度) (P513) Integral relation find V from : E 零

29、零勢勢點(diǎn)點(diǎn) P lEV d Differential relationfind from V: E VVE grad z V E y V E x V E zyx ; That is, Proof: E VVd V s d b a I II sEVV ba d VVVVVV ba d)d( VsEdd As figure, a,b are very near points, is the displacement from a to b. sd )ddd()(dkzj yi xkEjEiEsE zyx zEyExE zyx ddd z z V y y V x x V Vdddd z V E y

30、V E x V E zyx , Vk z V j y V i x V E )(i.e. Where, The operator of gradient in the right-coordinate 直角坐標(biāo)下直角坐標(biāo)下梯度算符梯度算符: k z j y i x )()()( )(grad,sok z V j y V i x V VVE （電場中某點(diǎn)場強(qiáng)沿某方向的分量等于（電場中某點(diǎn)場強(qiáng)沿某方向的分量等于電勢沿此方向空間變電勢沿此方向空間變 化率的負(fù)值化率的負(fù)值 或電場中某點(diǎn)場強(qiáng)等于該點(diǎn)的或電場中某點(diǎn)場強(qiáng)等于該點(diǎn)的電勢梯度的負(fù)值電勢梯度的負(fù)值）。）。 The component of in any direction is the negative of the rate of change of the electric potential with distance in that direction. E In SI system, t