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1、26THEELECTRICFIELDConceptualQuestions26.1.Atiny,positivetestchargewillbeplacedatthepointinspaceandtheforcewillbemeasured.FromtheforceFmeasurementandthechargetheelectricfieldwillbecalculatedusingE=.Thedirectionofthefieldwillbetheqsameasthedirectionofthef
2、orcesinceqispositive.26.2.26.3.E3421=>>EEE.Theelectricfieldstrengthislargerintheregionwherethefieldlinesareclosertogether(E34andE)andsmallerwherethefieldlinesarefartherapart.λλff(/)QLffLi26.4.(a)==.ButQQif,so==3λλii(/)QLiifLFf(b)F∝=λ,so3Fi(c)10timestheo
3、riginalamountofchargewouldgiveaconstantlinearchargedensity.Sotheamountofchargetoaddtotheoriginalis9timestheoriginalcharge.??12λ26.5.Fe==Ee??.Ifthechargedensityλisdoubled,thenthedistancerfromthewiremustalsobedoubled??4πE0rfortheforcetobethesame.Thusr=2cm
4、.26.6.Theelectricfieldatthecenteriszero.Wecanthinkofthestrawasbeingmadeupofmanyringsofpositivecharge.Atthecenteroftheringaddingallfieldvectorsgivesaresultantelectricfieldequaltozero,asshowninthefigurebelow.?Copyright2013PearsonEducation,Inc.Allrightsres
5、erved.Thismaterialisprotectedunderallcopyrightlawsastheycurrentlyexist.Noportionofthismaterialmaybereproduced,inanyformorbyanymeans,withoutpermissioninwritingfromthepublisher.26-126-2Chapter26AAiiηff(/)QA226.7.(a)Af==,so==3.163=103.1632ηii(/)QAAfFf(b)=1
6、.Thetotalchargeontheobjectdoesnotchangeasitshrinks.AnelectronveryfarfromtheobjectFiexperiencesitasapointcharge.Byveryfar,wemeanmuchfartherawaythanthesizeoftheobject.QQnCQQQQ1nC26.8.N1===228,andNN21===22=2==2.2A1πr1cmA2ππrrr211(2)4π4cmQEf126.9.(a)E=.IfQi
7、shalved,thenEisalsohalved.Thus=.4πEr2E20i(b)ThefieldoutsideasphereisthesameasthatofapointchargeQlocatedatthecenterofthesphere.SoiftheEfradiusofthespherechanges,thefieldremainsthesameoutsidethesphereatthedistancer=2R.So=1.Ei(c)Thefieldoutsideauniformlych
8、argedsphereisthesameasthatofapointchargeQlocatedatthecenterofthesphere.SincerisstillgreaterthanRafterbeinghalved,QEfEEfi==44?=r2E??i4πE0????226.10.DischargingtheballwillcausetherestoringforceFm=+(gqE)sin45°todecrease.Thereforethe
