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1、27GAUSS’SLAWConceptualQuestions27.1.No.Thecubelackssufficientsymmetrytodeducetheshapeofthefield.Inparticular,thecubelacksbothtranslationalandrotationalsymmetry.Itissymmetricunder90°rotationsbut,unlikeasphere,notsymmetricforarbitraryrotations.27.2.(a)Nonetchargeisenclosed.Thereisno
2、fluxthroughthebottomsurface,andthefluxintotheleft-handsideiscancelledbythefluxoutoftheright-handside.(b)Netpositivechargeisenclosed,sincethefluxisoutwardonallsurfaces.(c)Netnegativechargeisenclosedsincethefluxisinwardonallsurfaces.dddd27.3.FromEquation27.3,weknowthatΦ=⋅=squareEAEA
3、squaresquaresquaresquareandΦ=⋅=circleEAcirclecircleEAcirclecircle,wherethelastequalityholdsbecausetheelectricfieldisparalleltothesurfacenormal.WealsoknowEEAAsquare=>circleandsquarecircle,soΦsquareEAsquaresquareAsquare==>1⇒Φ>squareΦcircleΦcircleEAcirclecircleAcircle27.4.Φ=Φ12.Inthe
4、absenceofanetenclosedcharge,anyfluxintosurface1mustcomeoutofsurface2becausethetwosurfacesformasingleclosedsurface.27.5.(a)+q/e0(b)−q/e0(c)027.6.Φ=+A04/,qeΦ=−B04/,qeΦ=C0,Φ=+D03/,qeΦ=E027.7.Thechargeliesonthesurfaceofthehollowballoon.Point1:Staysthesame.Gauss’sLawwithQEin==0implies0
5、inbothsituations.Point2:Decreases.E>0initiallywhenthepointisoutsidetheballoonbutE=0attheendwhenthepointisinsidetheballoon.Point3:Staysthesame.Theelectricfieldoftheballoonlookslikethatofapointchargeatthegeometriccenteroftheballoonbothbeforeandaftertheballoonisblownup.27.8.Student1i
6、scorrect.ThetwospheresareGaussiansurfacesenclosingthesameamountofcharge,sothefluxthroughthetwosurfacesisequal,nomattertheirsize.Inthiscase,onecanseethattheareaincreasesasr2andtheelectricfieldstrengthdecreasesas1/r2sothefluxisthesamethroughspheresAandB.©Copyright2013PearsonEducatio
7、n,Inc.Allrightsreserved.Thismaterialisprotectedunderallcopyrightlawsastheycurrentlyexist.Noportionofthismaterialmaybereproduced,inanyformorbyanymeans,withoutpermissioninwritingfromthepublisher.27-127-2Chapter2727.9.Student2iscorrect.ThesphereandellipsoidareGaussiansurfacesenclosin
8、gthesameamountofcharge,sothefluxt
