教程大二上大物

1、Chap 20 Gausss LawMathematician & Physicist & astronomerElectric fluxSurface integralGausss lawClosed surface Gaussian surfaceKey terms:0 enclQAdE Gausss law is an more general relationship between electric charge and electric field. It is part of the key to use symmetry considerations to si

2、mplify electric-field calculations. And it gives us additional insight into the nature of electrostatic fields. Gausss law involves an integral of the electric field E at each point on a closed surface.20-1 Electric field lines1. Rule (P. 472): 1)The tangent to a field line is in the direction of th

3、e field;ANE By drawing few representative lines that follow the curvature mapped out by the arrows to visualize patterns in electric field. 2) The density of the lines measures the intensity of the field;EEANSome Typical Field Lines (P472):! !mind: electric field lines do not actually exist!2. Essen

4、ce of the Electric field lines (P 472)1) Electric field lines originate on positive charges (or come from infinity) and terminate on negative charges (or go on to infinity). They never originate or terminate on a charged-free point in finite space(不會(huì)在沒(méi)有電荷處中斷不會(huì)在沒(méi)有電荷處中斷).2) The field lines are never c

5、ross each other20-2 Electric Flux (P. 487)1. Definition: The number of field lines crossing a given area is called the flux through the area.1) In uniform fielde1EA2. Calculations of Electric FluxEAE if1AANnot if e cosEA22AEAn 2AEn AA 1AExample: The flux of the electric field (24 N/C)i + (30 N/C)j +

6、 (16 N/C)k through a 2.0 m2 portion of the yz plane is: A)32 N.m2/C B)34 N.m2/C C)42 N.m2/C D)48 N.m2/C E)60 N.m2/C22 A)REe Example:2 B)REe ER2) Non uniform fieldAEdAcosEedd AAeeAcosEAEddd EEdAn Divide the surface, and choose any small element dA. The field can be considered uniform over this tiny a

7、rea. The electric flux through the entire surface isAn dAA d3. Properties of Electric Flux 1) Electric flux is a scalar, but it can be positive or negative. The SI unit: Nm2/C, Vm.SSeAcosEAEdd Question:SSeAcosEAcosEdd Outward: positive ! 2) For a closed surface, we define the direction of A, or of d

8、A, to point outward from the enclosed volume. ;02ed, ;02ed, ;0d 2edA/E,En Leaving - positive EEntering - negative E;d 0,maxeeddAE,E/n 3) Electric flux for a closed surface is the net flux out of or into the volume.AAeAcosEAEdd Question: if there is no charges within a closed surface, what is the val

9、ue of electric flux? 0d 1SeAE 0 21e02e 1 2 3 0 3e20-3 Gausss law The precise relation between the electric flux through a closed surface (Gaussian surface) and the net charge Qencl enclosed within that surface is given by Gausss law (P. 489):0 enclQAdEWhere 0 is permittivity of free space.How to pro

10、ve the law? Here are the steps!1) suppose a point charge q is at the center of a sphere surface,11AedAcosE 1AdAE204rq0 q+qrrrqE42024 r 2) Any closed surface that enclosed the point charge.0122 qAdEAee1A2A3) Suppose q outside the closed surface02e1ee A+qA1A2q1q2q3q4q5521E.EEE A)E.EE(AEedd52100030201

11、qqqAE.AEAEddd5214) Suppose the surface encloses not just one point charge q but several chargesAdiienclAAeqAcosEAE01dd Notes:(1) E in Gausss law is the E on a closed surface, which we choose for our convenience. E is produced by all the charges within or out of the closed surface.(2) The flux only d

12、epends on the net charges inside. The charges out do not contribute to the flux, but to E over the Gaussian surface.But E over the surface must not be zero.00 (3) AiienclAdEq0 E,qAdEiienclA00 (4)But there must not be no charges within the closed surface.0 /iencliAE dAq(6) Gausss law, in principle, i

13、s valid in any electrostatic field, but only under the condition that the charge distribution has symmetric property we can use it to calculate intensity of the field.(5*)Charges are the source of electric field lines0EE Ans: CQuestion: Consider Gausss law: . Which of the following is true? A) must

14、be the electric field due to the enclosed charge.B) If then everywhere on the Gaussian surface.C) If the charge inside consists of an electric dipole, then the integral is zero. D) On the surface is everywhere parallel to .E) If a charge is placed outside the surface, then it cannot affect on the su

15、rface .0 enclQAdE E0 enclQEEAdEQuestion: Choose the INCORRECT statement: A) Gauss law can be derived from Coulombs law. B) Gauss law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface.C) Coulombs law c

16、an be derived from Gauss law and symmetry.D) Gauss law applies to a closed surface of any shape.E) According to Gauss law, if a closed surface encloses no charge, then the electric field must vanish everywhere on the surface. Ans: E20-4 Application of Gausss law: find E Symmetrical distribution of c

17、harges The magnitude of E on the gaussian surface is uniform or piecewise uniformUnder what conditions can we use Gausss law?What are the steps to apply Gausss law? Analyze symmetrical properties of the system (the charge distribution and the field). Draw a proper Gaussian surface (always be a spher

18、e or cylinder), the field points must be on the closed surface. Apply Gausss law on the surface, write the equation.Example: Solid sphere of charge. Positive charge Q is distributed uniformly throughout the volume of a sphere with radius R. Find the magnitude of the electric field at a point P a dis

19、tance r from the center of the sphere.Solution:qROrE must be equal magnitude over the sphere, and direct radially outward. Because of the spherical symmetry, we draw a cocentered sphere as shown in Fig.Rrowhen 2r4E 03034 rRQrEso Rrwhen 02qr4E 20r4qEso 3R4Q3 )3r4(3 )1(0 qROrROrE(Just like a point cha

20、rge!)Question : What if the charge were distributed on a spherical conductor？qROrRrowhen 0r4E2 0Eso Rrwhen 02qr4E 20r4qEso ROrEWhat if the conductor were a solid sphere?(Just like a point charge!)Example: Long uniform line of charge. Electric charge is distributed uniformly along an infinitely long,

21、 thin wire. The charge per unit length is . Find the electric field.Solution: Draw a closed cylinder with the wire along its axis as shown in Fig.rl Because of the cylindrical symmetry, the field is radially outward and depend only on the perpendicular distance, r, from the wire. SdE side2flat1flatS

22、dESdESdEsideSdE00rl2E 02 /lrlEr2E0 rl Apply Gausss law E is parallel to the flat ends of the cylinder cos900=0, it does not contribute to the flux. , AdEE is vertical to the cylinder , Ad/ECos00=1, andand E has the same magnitude over the cylinder. sidedSEChallenge Problem:For long uniform line of c

23、harger2E0 dPdrr？PE Finda dra rdrardE0022 addaddrdradEE02 dadlna02 Example: Infinite plane of charge. Find the electric field caused by a thin, flat, infinite sheet on which there is a uniform positive charge per unit area .Solution: Because of the plane symmetry, the field is direct perpendicular to

24、 the plane on both sides. Draw a closed cylinder whose axis is perpendicular to the plane and which extends through the plane as shown in Fig. SdE side2flat1flatSdESdESdEEA202 /AEA 02E 021flatflatSdESdE E is vertical to the flat ends of the cylinder cos00=1, and E must have the same magnitude on the

25、 two flat ends., AdEE is parallel to the cylinder , Ad/ECos900=0, andit does not contribute to the flux. AExample: Long solid cylinder of charge. A very long solid nonconducting cylinder of radius R0 and length L (R0R0) and (b) inside the cylinder (rR0). Do only for points far from the ends and for

26、which rL.0RSolution:0202 lRrlEE(a) Outside the cylinderrl0020 2Rr,rREE 0R022 lrrlEE(b) Inside the cylinder00 2Rr,rEE 2000000, 22, 2EERrRrrErrRQuestion : What if the charge were distributed on a long cylinder.000,20, crRrErR O Challenge Problem:P.501 prob.47orE31or32)(321rrooa3r1-r2aooFig501 aooP+=or

27、1po- r2porE3圖10-17 aooPP2r20-5 Charges on Conductors1) The electric field at every point within a conductor is zero.2) Any excess charge on a solid conductor is located entirely on its surface.3) If there is a cavity inside the conductor, and there is no charge in the cavity, In an electrostatic sit

28、uation (in which there is no net motion of charge), we get the following deductions:the excess charge on the conductor must be located only on its outer surface.q q q 4) If there is a cavity inside the conductor, and there is a charge q placed inside the cavity, there must be a total charge q on the cavity surface.5) Electrostatic shielding: If there is a cavity i