Physics 2102 Lecture 05: TUE 02 FEBWhat-Dat™ Flux!Gauss’ Law: General CaseExamplesFaraday’s Cage: Electric Field Inside Hollow Conductor is ZeroPowerPoint PresentationGauss’ Law: Cylindrical SymmetryGauss’ Law: Cylindrical SymmetryRecall Finite Line Example!Gauss’ Law: Insulating PlateInsulating and Conducting PlanesGauss’ Law: Spherical SymmetryElectric Fields With Spherical Symmetry: Shell TheoremSlide 14SummarySlide 16Physics 2102 Physics 2102 Lecture 05: TUE 02 FEBLecture 05: TUE 02 FEBGauss’ Law IIGauss’ Law IIPhysics 2102Jonathan DowlingQuickTime™ and a decompressorare needed to see this picture.Carl Friedrich Gauss1777 – 1855Flux Capacitor (Operational)What-DatWhat-Dat™™ Flux! Flux!QuickTime™ and a decompressorare needed to see this picture. ΦFlux=rEgrA = EA cosθQuickTime™ and a decompressorare needed to see this picture.Constant Area:AQuickTime™ and a decompressorare needed to see this picture. ΦFlux=rEgdrA—∫= E cosθ —∫dAChanging Area:dAEWhat-Dat™ is a RegisteredTrademark of Dowling Productions Inc.All Rights ReservedGauss’ Law: General Gauss’ Law: General CaseCase•Consider any ARBITRARY CLOSED surface S -- NOTE: this “Gaussian Surface” does NOT have to be a “real” physical object!•The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED!•The results of a complicated integral is a very simple formula: it avoids long calculations! € Φ ≡r E ⋅ dr A =±qε0Surface∫(One of Maxwell’s 4 equations!)SE±qExamplesExamples∫=⋅≡ΦSurface0εqAdErrWhat is FluxThrough Surfaces:S1 = S2 = S3 = S4 =+q/0–q/000Faraday’s Cage: Electric Faraday’s Cage: Electric Field Inside Hollow Field Inside Hollow Conductor is ZeroConductor is ZeroInside cavity is “shielded” from all external electric fields! “Faraday Cage effect”• Choose any arbitrary surface inside the metal• Since E = 0, flux = 0• Hence total charge enclosed = 0All charge goes on outer surface!Safe in the Plane!?QuickTime™ and a decompressorare needed to see this picture.Safe in the Car!?E=0QuickTime™ and aYUV420 codec decompressorare needed to see this picture.Gauss’ Law: Cylindrical Gauss’ Law: Cylindrical SymmetrySymmetry • Charge of qP=P10 C is uniformly spread over a line of length L = 1 m.• Use Gauss’ Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line.•Approximate as infinitely long line — E radiates outwards.•Choose cylindrical surface of radius R, length L co-axial with line of charge.R = 1 mmE = ?1 mLine of Charge: ?=?q/L Units: [C/m]Gauss’ Law: Cylindrical Gauss’ Law: Cylindrical SymmetrySymmetryRLEAE π2|||| ==Φ00ελεLq==ΦRkRRLLEλπελπελ222||00===•Approximate as infinitely long line — E radiates outwards.•Choose cylindrical surface of radius R, length L co-axial with line of charge.R = 1 mmE = ?1 mdA || E so cos??=?1Recall Finite Line Example!Recall Finite Line Example!∫−+=2/2/2/322)(LLyxadxakE λIf the Line Is Infinitely Long (L >> a) …2242LaaLk+=λ2/2/222LLaxaxak−⎥⎦⎤⎢⎣⎡+= λakLaLkEyλλ 222==QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture. € r E a?= RGauss’ Law: Insulating Gauss’ Law: Insulating Plate Plate •Infinite INSULATING plane with uniform charge density s•E is NORMAL to plane•Construct Gaussian box as shownAEA2 have, we,q law Gauss' Applying00=Φ=εσε02get wefield, electric for the Solvingεσ=EFor an insulator, E=/20, and for a conductor, E=/0. Surface Charge;=q/AUnits: [C/m2]Surface Charge;=q/AUnits: [C/m2]Insulating and Conducting Insulating and Conducting PlanesPlanesQ0022 εεσAQE ==Insulating Plate: Charge Distributed Homogeneously.002 εεσAQE ==Conducting Plate: Charge Distributed on the Outer Surfaces.Electric Field Inside a Conductor is ZERO!Q/2Gauss’ Law: Spherical Gauss’ Law: Spherical SymmetrySymmetry •Consider a POINT charge q & pretend that you don’t know Coulomb’s Law•Use Gauss’ Law to compute the electric field at a distance r from the charge•Use symmetry: –place spherical surface of radius R centered around the charge q–E has same magnitude anywhere on surface–E normal to surface0εq=ΦrqE24|||| rEAE π==Φ2204||rkqrqE ==πεElectric Fields With Electric Fields With Spherical Symmetry: Shell Spherical Symmetry: Shell TheoremTheorem-15C+10 CA spherical shell has a charge of +10C and a point charge of –15C at the center.What is the electric field produced OUTSIDE the shell?If the shell is conducting?Field Inside a Conductor is ZERO!If the shell is conducting?Field Inside a Conductor is ZERO!ErE=k(15C)/r2E=k(5C)/r2E=0And if the shell is insulating?Charged Shells Behave Like a Point Charge of Total Charge “Q” at the CenterOnce Outside the Last Shell!Charged Shells Behave Like a Point Charge of Total Charge “Q” at the CenterOnce Outside the Last Shell!ConductingQuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.Electric Fields With Insulating Electric Fields With Insulating SphereSpherer <Rqenclosed=QVenclosedVtotal⎛⎝⎜⎞⎠⎟=Q4πr3/ 34πR3/ 3⎛⎝⎜⎞⎠⎟=Qr3R3r <Rqenclosed=QVenclosedVtotal⎛⎝⎜⎞⎠⎟=Q4πr3/ 34πR3/ 3⎛⎝⎜⎞⎠⎟=Qr3R3r >Rqenclosed=Qr >Rqenclosed=QΦ =EA = qenclosed/ ε0Φ =EA = qenclosed/ ε0r <R→ E4πr2=Qr3R3/ε0r <R→ E4πr2=Qr3R3/ε0r >R→ E4πr2=Q /ε0r >R→ E4πr2=Q /ε0SummarySummary•Electric Flux: a Surface Integral (Vector Calculus!); Useful Visualization: Electric Flux Lines Like Wind Through a Window. •Gauss’ Law Provides a Very Direct Way to Compute the Electric Flux.QuickTime™ and a decompressorare needed to see this
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