Physics 2102Physics 2102Lecture 05: TUE 02 FEBLecture 05: TUE 02 FEBGaussGauss’’ Law IILaw IIPhysics 2102Jonathan DowlingCarl Friedrich Gauss1777 – 1855Flux Capacitor (Operational)What-DatWhat-Dat™™ Flux! Flux! !Flux=!Ei!A = EA cos"Constant Area:A !Flux=!Eid!A""= E cos#!""dAChanging Area:dAEWhat-Dat™ is a RegisteredTrademark of Dowling Productions Inc.All Rights ReservedGaussGauss’’ Law: General Case Law: General Case• Consider any ARBITRARYCLOSED surface S -- NOTE:this “Gaussian Surface” doesNOT have to be a “real”physical object!• The TOTAL ELECTRIC FLUXthrough S is proportional tothe TOTAL CHARGEENCLOSED!• The results of a complicatedintegral is a very simpleformula: it avoids longcalculations! ! "! E # d! A =±q$0Surface%(One of Maxwell’s 4 equations!)SE±qExamplesExamples!="#$Surface0%qAdE!!What isFluxThroughSurfaces:S1 =S2 =S3 =S4 =+q/ε0–q/ε000FaradayFaraday’’s Cage: Electric Fields Cage: Electric FieldInside Hollow Conductor is ZeroInside Hollow Conductor is ZeroInside cavity is “shielded”from all external electricfields! “Faraday Cageeffect”• Choose any arbitrary surfaceinside the metal• Since E = 0, flux = 0• Hence total charge enclosed = 0All charge goes on outer surface!Safe in thePlane!?Safe in the Car!?E=0GaussGauss’’ Law: Law: Cylindrical SymmetryCylindrical Symmetry• Charge of q!=!10 C isuniformly spread over a line oflength L = 1 m.• Use Gauss’ Law to computemagnitude of E at aperpendicular distance of 1 mmfrom the center of the line.•Approximate as infinitely longline — E radiates outwards.•Choose cylindrical surface ofradius R, length L co-axialwith line of charge.R = 1 mmE = ?1 mLine of Charge: λ!=! q/L Units: [C/m]GaussGauss’’ Law: Cylindrical Symmetry Law: Cylindrical SymmetryRLEAE!2|||| =="00!"!Lq==#RkRRLLE!"#!"#!222||00===•Approximate as infinitelylong line — E radiatesoutwards.•Choose cylindrical surfaceof radius R, length L co-axial with line of charge.R = 1 mmE = ?1 mdA || E so cos!θ!=!1Recall Finite Line Example!Recall Finite Line Example!!"+=2/2/2/322)(LLyxadxakE#If the Line Is Infinitely Long (L >> a) …2242LaaLk+=!2/2/222LLaxaxak!"#$%&'+=(akLaLkEy!!222== ! E a!= RGaussGauss’’ Law: Insulating Plate Law: Insulating Plate• Infinite INSULATING plane withuniform charge density s• E is NORMAL to plane• Construct Gaussian box as shownAEA2 have, we,q law Gauss' Applying00=!="#"02get wefield, electric for the Solving!"=EFor an insulator, E=σ/2ε0, and for a conductor, E=σ/ε0.Surface Charge;σ!=!q/AUnits: [C/m2]Insulating and Conducting PlanesInsulating and Conducting PlanesQ0022!!"AQE ==Insulating Plate: Charge Distributed Homogeneously.002!!"AQE ==Conducting Plate: Charge Distributed on the Outer Surfaces.Electric Field Inside a Conductor is ZERO!Q/2GaussGauss’’ Law: Spherical Symmetry Law: Spherical Symmetry• Consider a POINT charge q &pretend that you don’t knowCoulomb’s Law• Use Gauss’ Law to compute theelectric field at a distance r fromthe charge• Use symmetry:– place spherical surface of radiusR centered around the charge q– E has same magnitude anywhereon surface– E normal to surface0!q="rqE24|||| rEAE!=="2204||rkqrqE ==!"Electric Fields With SphericalElectric Fields With SphericalSymmetry: Shell TheoremSymmetry: Shell Theorem-15C+10 CA spherical shell has a charge of +10Cand a point charge of –15C at thecenter.What is the electric field producedOUTSIDE the shell?If the shell is conducting?Field Inside a Conductor is ZERO!ErE=k(15C)/r2E=k(5C)/r2E=0And if the shell is insulating?Charged ShellsBehave Like a Point Charge ofTotal Charge “Q” at the CenterOnce Outside the Last Shell!ConductingElectric Fields With Insulating SphereElectric Fields With Insulating Spherer < Rqenclosed= QVenclosedVtotal!"#$%&= Q4'r3/ 34'R3/ 3!"#$%&= Qr3R3r > Rqenclosed= Q! = EA = qenclosed/"0r < R ! E4"r2= Qr3R3/#0r > R ! E4"r2= Q /#0SummarySummary• Electric Flux: a Surface Integral (VectorCalculus!); Useful Visualization: Electric FluxLines Like Wind Through a Window.• Gauss’ Law Provides a Very Direct Way toCompute the Electric
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