Lecture 4 Lecture 4 --GaussGauss’’LawLawChapter 27 Chapter 27 --Thursday January 18thThursday January 18th•Review of Tuesday’s class•Electric field and field lines•Dipoles•Motion of charges in electric fields•Cathode rays and Millikan’s experiment•The Flux of a vector (in this case, electric) field•Gauss’ Law and Coulomb’s law•Applications of Gauss’ lawReading: pages 612 thru 621 (chapter 27) in HRKReading: pages 612 thru 621 (chapter 27) in HRKRead and understand the sample problemsRead and understand the sample problemsWebAssignWebAssign::set 1, due tonight at 11:59pmset 1, due tonight at 11:59pmset 2 will be due next Thur. check web siteset 2 will be due next Thur. check web siteGraded problems (Ch. 27) Graded problems (Ch. 27) ––Ex. 15, 18, 21; Prob. 2, 10, 18Ex. 15, 18, 21; Prob. 2, 10, 18Practice problems: Ch. 27 Practice problems: Ch. 27 --Ex. 3, 7, 17; Prob. 3Ex. 3, 7, 17; Prob. 302ˆqqKr=FrGElectric field Electric field --Chapter 26Chapter 26Analogy with gravitationAnalogy with gravitation++q0••Problem:Problem:Force depends on test chargeForce depends on test charge!!00orqq==FFE EGGG G••Definition:Definition:21ˆ4oqrπε=ErG••Same for all test charges.Same for all test charges.••Represents the electrostatic Represents the electrostatic influence (field) of the charge influence (field) of the charge qq. . q = −Qsource chargeElectric field Electric field --Chapter 26Chapter 260q=FEGGNewton’s law for electrostatics:There’s really no need for the “test charge”q=FEGGThis is the force on a charge q in an electric fieldEGUnits for Units for EEare N/C in this chapter are N/C in this chapter (later we shall use volts per meter)(later we shall use volts per meter)Electric field linesElectric field lines• The tangent to an electric field line at a point in space gives the direction of the electric field at that point.• The magnitude of the electric field at any point is proportional to the number of field lines per unit cross-sectional area perpendicular to the lines.• Electric field lines start on positive charges and end on negative charges (can also start/end at infinity). • The symmetry of the problem dictates the directions in which field lines radiate from charges.Electric field linesElectric field lines• The number of field lines radiating from a charge is proportional to the charge.Electric field linesElectric field linesThe Electric DipoleThe Electric DipoleDipole moment:On the median plane (along x axis):()314zopExdxπε=−>>On the dipole axis (along z axis):()312zopEzdzπε=>>pqd or q==pdGGMotion of charges in electric fieldsMotion of charges in electric fieldsqm==FE aGGGqm⇒=aEGGMotion of charges in electric fieldsMotion of charges in electric fieldsqm==FE aGGGqm⇒=aEGGProjectile motion (PHY 2060 or 2048)2100 00200yxyy xxyy vt at xx vtvv at vv−= + −==+ =Discovery of the electron: J. J. Thompson's experimentDiscovery of the electron: J. J. Thompson's experimentCathode ray tube Cathode ray tube --or CRTor CRTBasis for so much technology: TV, oscilloscope, mass spectrometeBasis for so much technology: TV, oscilloscope, mass spectrometer, etc...r, etc...Charged particles Charged particles evaporate from evaporate from heated cathode C. heated cathode C. Negative charges Negative charges accelerate in electric accelerate in electric field between the field between the cathode and anode A.cathode and anode A.Electric field betweenElectric field betweenD and E used to deflect beam.D and E used to deflect beam.Well focused beam of Well focused beam of negative particles passes negative particles passes through the slit B.through the slit B.Phosphorescent Phosphorescent screen used to screen used to detect the cathode detect the cathode ray.ray.Thompson also applied a Thompson also applied a magnetic field into the page, magnetic field into the page, which also deflected the ray.which also deflected the ray.In order to measure q/m it is essential to know the velocity of the beam.F = qE = maqaEm⇒=Millikan's oil drop experimentMillikan's oil drop experimenthttp://www.whfreeman.com/modphysics/PDF/3http://www.whfreeman.com/modphysics/PDF/3--1bw.pdf1bw.pdfMost dropsMost dropscarry a chargecarry a chargeElectric field appliedElectric field appliedbetween capacitor platesbetween capacitor platesto provide upwards forceto provide upwards forceon some charged oil dropson some charged oil dropsA source of XA source of X--rays was also used in the rays was also used in the experiment. Xexperiment. X--rays were found to knock rays were found to knock charges off from the oil drops, thereby charges off from the oil drops, thereby altering the total charge on the drops.altering the total charge on the drops.A Dipole in an Electric FieldA Dipole in an Electric FieldsinsinsinFdqdEpEτθθθ====×pEGGcosUpEθ=−=− ⋅pEGGForceForcerelated torelated togradient in the potential energygradient in the potential energyThings you must be able to do...Things you must be able to do...1. Calculate the force on a test charge produced by one or more discrete source charges. Similarly, calculate the electric field produced by a system of discrete charges.2. Be able to perform task #1 for continuous line charges, involving simple integrals.3. Be able to convert between total charge and charge density.4. Sketch electric field lines for different charge configurations.5. Know and understand the properties of dipoles.6. Be able to calculate the trajectories of point charges in uniform electric fields.7. Be able to calculate the torque and potential energy of an electric dipole in a uniform electric field.The flux of a vector fieldThe flux of a vector fieldConsider the velocity field of a flowing fluidConsider the velocity field of a flowing fluidIf the area A is flat, and perpendicular to the flow, then we define the flux Φ of the velocity field through the surface as follows: vGvAΦ=If v has dimensions of m/s, then the flux has dimensions of m3/s (volume flow rate).AcosθWhat if the area is not What if the area is not perpendicular to the flow? perpendicular to the flow? We project the area on to the plane perpendicular to the flow. Then,cosvAθΦ=The flux of a vector fieldThe flux of a vector fieldFor this extreme case (θ= 90o),0.Φ=What if the area is not What if the area is not perpendicular to the flow? perpendicular to the flow? The flux of a vector fieldThe flux of a vector fieldThe flux of a vector fieldThe flux of
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