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1、ErrataInstructor’sSolutionsManualIntroductiontoElectrodynamics,3rdedAuthor:DavidGriffithsDate:September1,2004•Page4,Prob.1.15(b):lastexpressionshouldready+2z+3x.•Page4,Prob.1.16:atthebeginning,insertthefollowingfigure•Page8,Prob.1.26:lastlineshouldreadFromProb.1.18:∇×
2、v=−6xzxˆ+2zyˆ+3z2ˆz⇒a∇·(∇×v)=∂(−6xz)+∂(2z)+∂(3z2)=−6z+6z=0.a∂x∂y∂z•Page8,Prob.1.27,inthedeterminantfor∇×(∇f),3rdrow,2ndcolumn:changey3toy2.•Page8,Prob.1.29,line2:thenumberintheboxshouldbe-12(insertminussign).•Page9,Prob.1.31,line2:change2x3to2z3;firstlineofpart(c):
3、insertcommabetweendxanddz.•Page12,Probl1.39,line5:removecommaaftercosθ.•Page13,Prob.1.42(c),lastline:insertˆzafter).•Page14,Prob.1.46(b):changertoa.•Page14,Prob.1.48,secondlineofJ:changetheupperlimitontherintegralfrom∞toR.Fixthelastlinetoread:R=4π−e−r+4πe−R=
4、4π−e−R+e−0+4πe−R=4π.0•Page15,Prob.1.49(a),line3:inthebox,changex2tox3.1•Page15,Prob.1.49(b),lastintegration“constant”shouldbel(x,z),notl(x,y).•Page17,Prob.1.53,firstexpressionin(4):insertθ,soda=rsinθdrdφθˆ.•Page17,Prob.1.55:Solutionshouldreadasfollows:Problem1.55(
5、1)x=z=0;dx=dz=0;y:0→1.v·dl=(yz2)dy=0;v·dl=0.(2)x=0;z=2−2y;dz=−2dy;y:1→0.v·dl=(yz2)dy+(3y+z)dz=y(2−2y)2dy−(3y+2−2y)2dy;0432032y4yy14v·dl=2(2y−4y+y−2)dy=2−+−2y=.232311(3)x=y=0;dx=dy=0;z:2→0.v·dl=(3y+z)dz=zdz.00z2v·dl=zdz==−2.222Total:v·dl=0+14−2=8.33Me
6、anwhile,Stokes’thereomsaysv·dl=(∇×v)·da.Hereda=dydzxˆ,soallweneedis(∇×v)=∂(3y+z)−∂(yz2)=3−2yz.Thereforex∂y∂z12−2y(∇×v)·da=(3−2yz)dydz=(3−2yz)dzdy00112132=3(2−2y)−2y(2−2y)dy=(−4y+8y−10y+6)dy0210=−y4+8y3−5y2+6y=−1+8−5+6=8.3033•Page18,Prob.1.56:change(3)and(
7、4)toreadasfollows:(3)φ=π;rsinθ=y=1,sor=1,dr=−1cosθdθ,θ:π→θ≡2sinθsin2θ20tan−1(1).2cos2θcosθcosθsinθv·dl=rcos2θ(dr)−(rcosθsinθ)(rdθ)=−dθ−dθsinθsin2θsin2θcos3θcosθcosθcos2θ+sin2θcosθ=−+dθ=−dθ=−dθ.sin3θsinθsinθsin2θsin3θThereforeθ0θ0cosθ11151v·dl=−dθ==−=−=
8、2.sin3θ2sin2θ2·(1/5)2·(1)22π/2π/22√(4)θ=θ,φ=π;r:5→0.v·dl=rcos2θ(dr)=4rdr.0250044r245v·dl=rdr==−·=−2.552√52√55Total:3πv·dl=0++2−2=3π.22•
