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1、CHAPTER44.1.ThevalueofEatP(ρ=2,φ=40◦,z=3)isgivenasE=100a−200a+300aV/m.ρφzDeterminetheincrementalworkrequiredtomovea20µCchargeadistanceof6µm:a)inthedirectionofaρ:TheincrementalworkisgivenbydW=−qE·dL,whereinthiscase,dL=dρa=6×10−6a.ThusρρdW=−(20×10−6C)(100V/m)(6×10−6m)=−12×10−9J=−12nJb)in
2、thedirectionofa:InthiscasedL=2dφa=6×10−6a,andsoφφφdW=−(20×10−6)(−200)(6×10−6)=2.4×10−8J=24nJc)inthedirectionofa:Here,dL=dza=6×10−6a,andsozzzdW=−(20×10−6)(300)(6×10−6)=−3.6×10−8J=−36nJd)inthedirectionofE:Here,dL=6×10−6a,whereE100aρ−200aφ+300azaE=[1002+2002+3002]1/2=0.267aρ−0.535aφ+0.802
3、azThusdW=−(20×10−6)[100a−200a+300a]·[0.267a−0.535a+0.802a](6×10−6)ρφzρφz=−44.9nJe)InthedirectionofG=2a−3a+4a:Inthiscase,dL=6×10−6a,wherexyzG2ax−3ay+4azaG=[22+32+42]1/2=0.371ax−0.557ay+0.743azSonowdW=−(20×10−6)[100a−200a+300a]·[0.371a−0.557a+0.743a](6×10−6)ρφzxyz=−(20×10−6)[37.1(a·a)−55
4、.7(a·a)−74.2(a·a)+111.4(a·a)ρxρyφxφy+222.9](6×10−6)where,atP,(a·a)=(a·a)=cos(40◦)=0.766,(a·a)=sin(40◦)=0.643,andρxφyρy(a·a)=−sin(40◦)=−0.643.SubstitutingtheseresultsinφxdW=−(20×10−6)[28.4−35.8+47.7+85.3+222.9](6×10−6)=−41.8nJ404.2.Apositivepointchargeofmagnitudeq1liesattheorigin.Derive
5、anexpressionfortheincrementalworkdoneinmovingasecondpointchargeq2throughadistancedxfromthestartingposition(x,y,z),inthedirectionof−ax:TheincrementalworkisgivenbydW=−q2E12·dLwhereE12istheelectricfieldarisingfromq1evaluatedatthelocationofq2,andwheredL=−dxax.Takingthelocationofq2atspherica
6、lcoordinates(r,θ,φ),wewrite:−q2q1dW=ar·(−dx)ax4π≤0r2wherer2=x2+y2+z2,andwherea·a=sinθcosφ.Sorxpq2q1x2+y2xq2q1xdxdW=ppdx=4π≤0(x2+y2+z2)x2+y2+z2x2+y24π≤0(x2+y2+z2)3/2
7、{z}
8、{z}sinθcosφ4.3.IfE=120aρV/m,findtheincrementalamountofworkdoneinmovinga50µmchargeadistanceof2mmfrom:a)P(1,2,3)towardQ(
9、2,1,4):ThevectoralongthisdirectionwillbeQ−P=(1,−1,1)√fromwhichaPQ=[ax−ay+az]/3.Wenowwrite∑∏−6(ax−ay+az−3dW=−qE·dL=−(50×10)120aρ·√(2×10)3−61−3=−(50×10)(120)[(aρ·ax)−(aρ·ay)]√(2×10)3AtP,φ=tan−1(2/1)=63.4◦.Thus(a·a)=cos(63.4)=0.447and(a·a)=ρxρysin(63.4)=0.894.Substitutingthese,weobtaind