FundamentalsofPhysicsHalliday&Resnic東海大學物理系第頁，共NUMPAGES13頁【CH23】FindingtheElectricFieldII1,3,5,7,15,19,21,23,29,33,35,37,41,45,49,51【Problem23-1】ThesquaresurfaceshowninFig.23-26measures3.2mmoneachside.Itisimmersedinauniformelectricfieldwithmagnitudeandwithfieldlinesatanangleofwithanormaltothesurface,asshown.Takethatnormaltobedirected“outward,”asthoughthesurfacewereonefaceofabox.Calculatetheelectricfluxthroughthesurface.圖23-26中，一邊長3.2mm的正方形平面，被放置於一強度為的均勻電場中，平面的法向量與電力線的夾角。將這個平面當作箱子的其中一個面，法向量的方向指向外。計算通過此平面的電通量。（圖23-26）<解>：Thevectorareaandtheelectricfieldareshownonthediagrambelow.Theanglebetweenthemis180°–35°=145°,sotheelectricfluxthroughtheareais【Problem23-3】ThecubeinFig.23-27hasedgelength1.40mandisorientedasshowninaregionofuniformelectricfield.Findtheelectricfluxthroughtherightfaceiftheelectricfield,innewtonspercoulomb,isgivenby(a),(b),and(c).(d)Whatisxthetotalfluxthroughthecubeforeachfield?圖Fig.23-27中，一邊長1.40m的正立方體，被放置在一均勻電場中，擺放方式如圖所示。請求出右側表面的電通量，當電場分別為：(a)N∕C；(b)N∕C；(c)N∕C。(d)請分別求出在上述三個不同的電場條件下，通過此立方體的總電通量？（圖23-27）<解>：Weuse,where.(a)(b)(c).(d)Thetotalfluxofauniformfieldthroughaclosedsurfaceisalwayszero.【Problem23-5】ApointchargeofisatthecenterofacubicalGaussiansurface55cmonedge.Whatisthenetelectricfluxthroughthesurface?一帶電量為的點電荷，位於一正立方體的高斯面中心，已知正立方體的邊長為55cm。請問通過此高斯面的淨電通量為？<解>：WeuseGauss’law:,whereisthetotalfluxthroughthecubesurfaceandqisthenetchargeinsidethecube.Thus,【Problem23-7】InFig.23-29,aprotonisadistancedirectlyabovethecenterofasquareofside.Whatisthemagnitudeoftheelectricfluxthroughthesquare?(Hint:Thinkofthesquareasonefaceofacubewithedge.)圖23-29中，有一邊長為的正方形，一個質子為於正方形中心正上方處。請問通過正方形的電通量大小為？（提示：將正方形視作邊長等於之立方體的其中一個面。）（圖23-29）<解>：Toexploitthesymmetryofthesituation,weimagineaclosedGaussiansurfaceintheshapeofacube,ofedgelengthd,withaprotonofchargesituatedattheinsidecenterofthecube.Thecubehassixfaces,andweexpectanequalamountoffluxthrougheachface.Thetotalamountoffluxisnet=q/0,andweconcludethatthefluxthroughthesquareisone-sixth