-field in regions I, II, and III. Todo so, you will need to first find the electricfield due to the left plate with charge 2Qateach position.Then due the same for theright plate with charge−Q. For both cases,assume the charge distributes uniformly overthe charged plate.Answer options are in the following order:EI, EII, EIII.1.Q2Aǫ0(−ˆx),3Q2Aǫ0ˆx,Q2Aǫ0ˆxcorrect2.3Q2Aǫ0(−ˆx),Q2Aǫ0ˆx,3Q2Aǫ0ˆx3.QAǫ0(−ˆx),3Q2Aǫ0ˆx,Q2Aǫ0(−ˆx)4.Q2Aǫ0(−ˆx),3Q4Aǫ0ˆx,Q4Aǫ0(−ˆx)Explanation:The left plate with charge 2Quniformlydistributed on its plates give an electric fieldof a large uniformly charged disk or,E2QI=−QAǫ0ˆx, E2QII=QAǫ0ˆx, E2QIII=QAǫ0ˆx .The right plate with charge−Quniformlydistributed on its plates give an electric fieldof a large uniformly charged disk or,E−QI=Q2Aǫ0ˆx, E−QII=Q2Aǫ0ˆx, E−QIII=−Q2Aǫ0ˆx .The fields in the three regions are given bythe superposition of theE2QandE−Qfields:EI=−Q2Aǫ0ˆx, EII=3Q2Aǫ0ˆx, EIII=Q2Aǫ0ˆx .008(part2of3)10.0pointsUsing Gauss’ law and the fields you deter-mined in Part 1, calculate the surface chargedensities on the inner and outer surfaces ofeach plate. The answer options are presentedin the form: (σL, σR),(σ′L, σ′R).Hint: Con-sider such Gaussian surfaces asScandSdwhich have one end inside a conducting plate.1.parenleftbiggQA,QAparenrightbigg,parenleftbigg−Q2A,−Q2Aparenrightbigg2.parenleftbiggQ2A,3Q2Aparenrightbigg,parenleftbigg−3Q2A,Q2Aparenrightbiggcorrect3.parenleftbiggQ2A,3Q2Aparenrightbigg,parenleftbigg−3Q2A,−Q2Aparenrightbigg4.parenleftbiggQ2,3QAparenrightbigg,parenleftbigg−3QA,QAparenrightbiggExplanation:We use the Gaussian surfacesScandSdto illustrate the relationship between the sur-face charge densityσof a conducting surfaceand theE-field perpendicular to that surface.Evaluating Gauss’ law over surfaceSc(lettingit’s cross-sectional area bea),contintegraldisplayvectorE•dvectorA=Qencǫ0ǫ0EIIa=σRaσR=ǫ0EII=32QA.Repeating the procedure usingSd, one findsthatσ′L=−σR=−32QA.Similar Gauss’ law calculations on the outersurfaces leads to the final answer:parenleftbiggQ2A,3Q2Aparenrightbigg,parenleftbigg−3Q2A,Q2Aparenrightbigg.009(part3of3)10.0pointsNow consider a similar physical configura-tion for which different information is known,shown in Fig.B. This time, you know that
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forsythe (acf2464) – HW8-c – li – (55020)6the plates are equally but oppositely charged,separated by a distanced, and a potential dif-ferenceVexists between them.Determinethe surface charge density on the inner sur-face of the right plate in terms ofV , dandphysical constants.
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