Thus e q in 4 π ? r 2 q 4 π ? r 2 where r r 1 this

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ThusE=qin4π ǫ0r2=Q4π ǫ0r2,whereR < r .(1)This is the familiar electric field due to a pointcharge that was used to develop Coulomb’slaw.The electric field forr < Rwith thesphere conducting:In the region insidethe conducting sphere, we select a sphericalGaussian surfacer < R, concentric with theconducting sphere.To apply Gauss’s lawin this situation, we realize that there is nocharge within the Gaussian surface (qin= 0),which implies thatE= 0,wherer < R .(2)rR01rE
homework 03 – KIM, JI – Due: Feb 3 2008, 4:00 am7Consider the expressions for the electric po-tential for outside and inside the conductingsphere.The electric potential forR < rwiththe conducting and/or uniformly non-conducting sphere:Previously we foundthat the magnitude of the electric field outsidea charged sphere of radiusRisE=kQr2,whereR < r ,where the field is directed radially outwardwhenQis positive.In this case, to obtain the electric potentialat an exterior point, we use the definition forelectric potential:V=-integraldisplayrE dr=-k Qintegraldisplayrdrr2=kQr,whereR < r .(4)Note:This result is identical to the expressionfor the electric potential due to a point charge.The electric potential forr < Rwiththe conducting sphere:In the region insidethe conducting sphere, the electric fieldE=0.Therefore the electric potential everywhereinside the conducting sphere is constant; thatisV=V(R)= constant,whereR < r .(5)S.rR01rVQuestion 11, chap 25, sect 2.part 2 of 210 pointsWhich diagram describes the electric po-tentialvsradial distance [V(r) function] for auniformly charged non-conductingsphere?1.L2.Gcorrect3.P4.M5.SExplanation:The electric potential forR < rwiththe uniformly non-conducting sphere:In the region outside the uniformly chargednon-conducting sphere, we have the same con-ditions as for the conducting sphere when ap-plying the definition for the electric potential,soV=-integraldisplayrE dr=-k Qintegraldisplayrdrr2=kQr,whereR < r .(4)The electric potential forr < Rwiththe sphere uniformly non-conducting:Because the potential must be continuous atr=R ,we can use this expression to obtainthe potential at the surface of the sphere;i.e.,the potential at a point on the conductingsphere isV=kQr.The electric field forr < Rwith theuniformly non-conducting sphere:Inthis case we select a spherical gaussian sur-face at a radiusrwherer < R, concentricwith the uniformly charged non-conductingsphere.Let us denote the volume of thissphere byV.To apply Gauss’s law in thissituation, it is important to recognize that thechargeqinwithin the gaussian surface of thevolumeVis less thanQ. Using the volumecharge densityρQV,we calculateqin:qin=ρ V=ρparenleftbigg43π r3parenrightbigg.
homework 03 – KIM, JI – Due: Feb 3 2008, 4:00 am8By symmetry, the magnitude of the electricfield is constant everywhere on the sphericalgaussian surface and is normal to the surfaceat each point. Therefore, Gauss’s law in theregionr < RgivescontintegraldisplayE dA=EcontintegraldisplaydA=E(4π r2)=qinǫ0

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