大学物理下课件23-Gass'law

1、大学物理下课件23-GasslawFluxExample of flux: A uniform air stream flowing through a square area. Flux is defined as volume flow rate (volume of air flow through the loop per unit time):AvAv and between angle :area vector :vector velocity :cosvAAvFlux of an Electric FieldThe flux of electric field: the elec

2、tric field passing through a certain area A. Considering a small area A, the flux passing through it isEA Note: can be positive, negative or zero depends on the direction of the vectors. Area 3: The total flux through this surface is equal to positive.Examples in the figure:Area 1: The total flux on

3、 the left side of the Gaussian surface is negative .Area 2: The total flux through this surface is zero.Gaussian surface: an imaginary surfaceThe electric flux through a Gaussian surface: proportional to the net number of field lines passing through that surface.For discrete surface:AEFor continuous

4、 surface:AdEExample: Figure shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform electric field E, with the cylinder axis parallel to the field. Find the flux of the electric field through the closed surface of the cylinder.EaTotal flux = flux through surface a + flux

5、 through surface b + flux through surface cbc000abcE dA cos180 + E dA cos 0 + E dA cos 90 - E (caps area) + E (caps area) + 0 = 0 900abc-EdA + E dA + 0 dA Solution:dAEEdAdAE180000cbaAdEAdEAdEExample: Find the flux of spherical Gaussian surface of radius r which encloses a positive point charge +q at

6、 its center.+qSolution:At every point on the surface: = 0, E = constant2201 A44qErrNote: It holds for any Gaussian surface that encloses the charge q and for any location of q inside AdAEA2A1+E on surfaceE1 E2Surface 1Surface 2 Total electric field lines crossing the whole surfacesameEAdAEAdE0220441

7、qrrqEA21AAGauss LawIf there are positive charges inside the surface, which means field lines are created inside the Gaussian surface.If there are negative charges inside the surface, which means field lines are terminated inside the Gaussian surface.Gauss law:charge enclosed by the hypotheticalclose

8、d surface(Gaussian surface) A.Gauss law relates the electric fields at points on a closed Gaussian surface to the net charge enclosed by that surfaceAencqAdE0The electric field is outward for all points on this closed surface.The flux and charge inside are both positive.The electric field is inward

9、for all points on the surface.The flux and charge inside are both negative.Net charge inside this closed surface is zero.Number of the electric field lines entering the surface = number of lines leaving it.No Charge inside the surface.The electric field lines enter the surface and also leave it.Illu

10、stration of Gauss lawGauss law and Coulombs lawBoth describe the electric field distribution around a system of charges. Consider a spherical Gaussian surface around a point charge q, apply the Gauss lawNote that E and dA has the same direction on every point of the surface. because of symmetry, EdA

11、 is the same on every point and the integral becomes:It gives the Coulombs law at the end:dAGaussian spherical surface of radius rEqPoint chargeThe Coulombs law can be derived from the Gausss law:encqAdE0encqEdAAdE00Coulombs law:The Gausss law:encqAdE02041rqEqrE2042041rqE Electrostatic equilibrium C

12、onductorinside E = 0 ConductorIf inside E 0, free electrons will move. Not electrostatic equilibriumElectrons always present in any conductorEF In electrostatic equilibrium, the electric field inside a conductor is zeroA charged isolated conductorExcess charge of a conductor In electrostatic equilib

13、riumConductor inside E = 0Gaussian s u r f a c e inside the conductorSince the electric field inside the conductor = 0Flux through any Gaussian surface inside the conductor = 0 Net charge inside the conductor = 0In electrostatic equilibrium, all excess charge on a conductor is entirely on the conduc

14、tors outer surface.+ An isolated Conductor with a CavitycavityGaussian surface0intEint0ernalENo net charge on the cavity walls, all the excess charge remains on the outer surface of the conductor.The cavity doesnt change the distribution of the excess charges.00inAE dAqNo electric charge q in the ca

15、vityAn electric charge q in the cavityint0ernalEinqq00inAE dAqqThe electric charge on the inner surface of the conductor is:Gaussian surface0intE+qThe external electric field at point p just outside the conductors surface:The External Electric Field0e x tEcharge per unit area on the conductors surfa

16、ce at point pEAintAernalcurvedexternalE dAE dAE dAE dA00e x tEA0A0extEWhile there is no field inside a conductor, the charges on the surface generate an electric field outside the surface.1.Consider a flat surface with uniform charge density .2.Draw a cylindrical Gaussian surface as shown. 3.The net

17、 charge enclosed by the Gaussian surface is A and the flux through the outside surface is EA4.Note that the field on the left is zero and hence there is no flux on the left surface.Higher charge densityIn general, electric field is very difficult to calculate.Charge distribution is not uniformLower

18、charge density ConductorElectric field is very easy to calculate.Uniform charge distribution ConductorsphereElectric field from non-uniform charge distribution-q-+Example: A negative point charge q = -5 C is located at a distance R/2 to the center of an electrically neutral spherical metal shell, fi

19、nd the distribution of charges on the shell.The distribution of charge on the inner wall is skewed because the negative charge is off-center.outer wall:outinqqqThe distribution of negative charge on the outer wall is uniform. Why?Solution:Inside the conductor, E = 0.Since there is no field inside th

20、e conductor, there is no flux through the Gaussian surface.By Gauss law, the net charge enclosed by the Gaussian surface must be zero. 0 = qenc = 0So the induced charges on the inner surface = +5C.Application of Gauss LawUsing Gauss law to calculate the electric field produced by some symmetric arra

21、ngements of charge.The Problem is how to choose a proper Gaussian surface.+1.The Gaussian surface should match the symmetry of the charge.2. = constant and E = constant at every point, we can put E and out of the integral. 3.The summation of A must be very simple.dAEAdEcosdAEcosrhApplying Gauss Law:

22、 Cylindrical Symmetry+r Problem: An infinite long cylindrical plastic rod with a uniform positive linear charge density . Find the electric field at a distance r from the axis of the rod.Solution:1.Choose a cylindrical Gaussian surface which can match the cylindrical symmetry of the problem.2.Find f

23、rom the cylindrical symmetry that E must have the same magnitude and ( 0) must be directed radically outward at any point on the cylindrical surface.Flux through the end caps: =00 (infinite line of charge)2ErDirection: radially outward when 0.hqAdEAenc0rhEdAEAdEcylindercylinder2000Applying Gauss Law

24、 : Planar Symmetry of Nonconducting SheetProblem: A thin, infinite nonconducting sheet with a uniform surface charge density + on one side. Find the electric field E at a distance r in front of the sheet. Solution:1.From symmetry, E must be perpendicular to the sheet and have the same magnitude at t

25、he same distance r.+r2.Choose a cylindrical Gaussian surface and it pierces to the sheet perpendicularly.0encAE dAq0(0)EAEAA02EDirection: perpendicular to the sheet.It is a uniform electric field.It also holds for a finite sheet, at points close to the sheet and not too near its edges.EAoEMagnitude:

26、Put two thin infinite plates with uniform surface charge density on both faces to be close to each other.Problem: if the plates are nonconducting, find E .+-1.The distribution of charge wont change since the excess charge on an insulator cant move.The net electric field is the vector sum of the fiel

27、ds due to four infinite plates.0000()02222LE Left to the plates:Between the plates:Right to the plates:0000022222BE0000()()02222RE (take rightward as positive direction)Two Conducting PlatesProblem: if the plates are conducting, Find EThe excess charge of the two conductors can move and will move onto the inner faces because the charge attract each other.0022022LE Left to the plates:Between the plates:Right to the plates:00022222BE0022()022RE be directed rightwardThe net electric field is the