From last time…PowerPoint PresentationElectric FluxWhy perpendicular component?Total fluxQuick quizFlux through spherical surfaceSlide 8Gauss’ lawUsing Gauss’ lawField outside uniformly-charged sphereQuick QuizE-field from line of chargeSlide 14Field from infinite plane of chargeProperties of Conductor in Electrostatic EquilibriumInduced dipoleConductors: charge on surface onlyField’s Magnitude and DirectionSep. 27, 2007 Physics 208 Lecture 8 1From last time…This Friday’s honors lecture:Biological Electric FieldsDr. J. Meisel, Dept. of Zoology, UWContinuous charge distributionsy++++++++++++++++++++ € dr E Motion of charged particlesSep. 27, 2007 Physics 208 Lecture 8 2Exam 1 Wed. Oct. 3, 5:30-7 pm, 2103 Ch (here)Covers Chap. 21.6, 22-27+ lecture, lab, discussion, HWSample exams on website.Solutions on website.HW4 (assigned today) covers exam material.Sep. 27, 2007 Physics 208 Lecture 8 3Electric FluxElectric flux E through a surface: (component of E-field surface) X (surface area)Proportional to # E- field lines penetrating surfaceSep. 27, 2007 Physics 208 Lecture 8 4Why perpendicular component?Suppose surface make angle surface normalE = EA cos E =0 if E parallel AE = EA (max) if E AFlux SI units are N·m2/C € r E = E||ˆ s + E⊥ˆ n € ˆ n € ˆ s € r A = Aˆ n Component || surfaceComponent surfaceOnly component‘goes through’ surface € ΦE=r E •r ASep. 27, 2007 Physics 208 Lecture 8 5Total fluxE not constantadd up small areas where it is constant Surface not flatadd up small areas where it is ~ flat € δΦEi= EiδAicosθ =r E i• δr A iAdd them all up: € ΦE=r E • dr A surface∫Sep. 27, 2007 Physics 208 Lecture 8 6qQuick quizTwo spheres of radius R and 2R are centered on the positive charges of the same value q. The flux through these spheres compare asA) Flux(R)=Flux(2R)B) Flux(R)=2Flux(2R)C) Flux(R)=(1/2)Flux(2R)D) Flux(R)=4Flux(2R)E) Flux(R)=(1/4)Flux(2R)qR2RSep. 27, 2007 Physics 208 Lecture 8 7Flux through spherical surfaceClosed spherical surface, positive point charge at centerE-field surface, directed outwardE-field magnitude on sphere: Net flux through surface:E constant, everywhere surface€ E =14πεoqR2R € ΦE=r E • dr A ∫= EdA∫= E dA∫= EA =14πεoqR2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟4πR2( )= q /εoSep. 27, 2007 Physics 208 Lecture 8 8qQuick quizTwo spheres of the same size enclose different positive charges, q and 2q. The flux through these spheres compare asA) Flux(A)=Flux(B)B) Flux(A)=2Flux(B)C) Flux(A)=(1/2)Flux(B)D) Flux(A)=4Flux(B)E) Flux(A)=(1/4)Flux(B)2qABSep. 27, 2007 Physics 208 Lecture 8 9Gauss’ lawnet electric flux through closed surface = charge enclosed / € ΦE=r E • dr A ∫=QenclosedεoSep. 27, 2007 Physics 208 Lecture 8 10Using Gauss’ lawUse to find E-field from known charge distribution.Step 1: Determine direction of E-fieldStep 2: Make up an imaginary closed ‘Gaussian’ surfaceMake sure E-field at all points on surface either perpendicular to surfaceor parallel to surfaceMake sure E-field has same value whenever perpendicular to surfaceStep 3: find E-field on Gaussian surface from charge enclosed.Use Gauss’ law: electric flux thru surface = charge enclosed / o € Φ =r E • dr A ∫= Qin/εoSep. 27, 2007 Physics 208 Lecture 8 11Field outside uniformly-charged sphereField direction: radially out from chargeGaussian surface:Sphere of radius rSurface area where Value of on this area: Flux thru Gaussian surface: Charge enclosed: € r E • dr A ≠ 0 :€ 4πr2 € r E • dr A € E€ Q€ E 4πr2Gauss’ law: € E 4πr2= Q /εo⇒ E =14πεoQr2Sep. 27, 2007 Physics 208 Lecture 8 12Quick QuizWhich Gaussian surface could be used to calculate E-field from infinitely long line of charge?None oftheseA. B. C. D.Sep. 27, 2007 Physics 208 Lecture 8 13E-field from line of chargeField direction: radially out from line chargeGaussian surface:Cylinder of radius rArea where Value of on this area: Flux thru Gaussian surface: Charge enclosed: € r E • dr A ≠ 0 :€ 2πrL € r E • dr A € E€ λL€ E2πrLGauss’ law: € E2πrL = λL /εo⇒ E =12πεoλrLinear charge density C/mSep. 27, 2007 Physics 208 Lecture 8 14Quick QuizWhich Gaussian surface could be used to calculate E-field from infinite sheet of charge?Any oftheseA. B. C. D.Sep. 27, 2007 Physics 208 Lecture 8 15Field from infinite plane of chargeField direction: perpendicular to planeGaussian surface:Cylinder of radius rArea where Value of on this area: Flux thru Gaussian surface: Charge enclosed: € r E • dr A ≠ 0 :€ 2πr2 € r E • dr A € E€ ηπr2€ E2πr2Gauss’ law: € E2πr2= η πr2/εo⇒ E =η2εoSurface charge C/m2Sep. 27, 2007 Physics 208 Lecture 8 16Properties of Conductor in Electrostatic EquilibriumIn a conductor in electrostatic equilibrium there is no net motion of chargeProperty 1: E=0 everywhere inside the conductorConductor slab in an external field E: if E 0 free electrons would be accelerated These electrons would not be in equilibriumWhen the external field is applied, the electrons redistribute until they generate a field in the conductor that exactly cancels the applied field.Etot = E+Ein= 0EinEtot =0Sep. 27, 2007 Physics 208 Lecture 8 17Induced dipoleWhat charge is induced on sphere to make zero electric field?+Sep. 27, 2007 Physics 208 Lecture 8 18Conductors: charge on surface onlyChoose a gaussian surface inside (as close to the surface as desired)There is no net flux through the gaussian surface (since E=0)Any net charge must reside on the surface (cannot be inside!)E=0Sep. 27, 2007 Physics 208 Lecture 8 19Field’s Magnitude and DirectionE-field always surface:Parallel component of E would put force on chargesCharges would accelerateThis is not equilibriumApply Gauss’s lawField
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