Physics 2102 Lecture: 07 WED 28 JANGauss’ Law: General CaseExamplesGauss’ Law: Cylindrical SymmetryGauss’ Law: Cylindrical SymmetryRecall Finite Line Example!Gauss’ Law: Insulating PlateInsulating and Conducting PlanesGauss’ Law: Spherical SymmetryElectric Fields With Spherical Symmetry: Shell TheoremPowerPoint PresentationSummarySlide 13Physics 2102 Physics 2102 Lecture: 07 WED 28 JANLecture: 07 WED 28 JANGauss’ Law IIIGauss’ Law IIIVersion: 1/22/07Physics 2102Jonathan DowlingQuickTime™ and a decompressorare needed to see this picture.Carl Friedrich Gauss1777 – 1855Flux Capacitor (Operational)Gauss’ Law: General Gauss’ Law: General CaseCase•Consider any ARBITRARY CLOSED surface S -- NOTE: this 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∫S(One of Maxwell’s 4 equations!)ExamplesExamples∫=⋅≡ΦSurface0εqAdErrGauss’ Law: Cylindrical Gauss’ Law: Cylindrical SymmetrySymmetry • Charge of 10 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 mGauss’ 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 mRecall 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 a2= 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.Insulating and Conducting Insulating and Conducting PlanesPlanesQ0022 εεσAQE ==Insulating Plate: Charge Distributed Homogeneously.002 εεσAQE ==Conducting Plate: Charge Distributed on the Outer Surfaces.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!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 Spherical Electric Fields With Spherical Symmetry: InsulatingSymmetry: InsulatingSummarySummary•Electric Flux: a Surface Integral (Vector Calculus!); Useful Visualization: Electric Flux Lines Caught by the Net on the Surface. •Gauss’ Law Provides a Very Direct Way to Compute the Electric Flux.QuickTime™ and a decompressorare needed to see this
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