Solution Manual for Classical Electromagnetism, 1st Edition

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’s Solutions Manualto accompanyClassical ElectromagnetismJerrold FranklinDecember 11, 2005

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iPrefaceThis instructor’s Solutions Manual contains my worked out solutions for all the problemsin the text “Classical Electromagnetism”. I have tried to be complete in each answer, andto include all the steps leading to a sraightforward solution.As much as possible I havewritten down the solutions as I would in doing them on the blackboard, including enoughexplanation to make each step reasonable.When a problem requires a particular equation from the book, I have included a referenceenclosed in square brackets, as, for instance, [Eq. (1.24)].References to equations in thesolutions manual are given without the square brackets.If I use an equation without areference to the text, that is an indication that I have expected the students to know thatequation without going back to the text. References to Sections in the text are also given insquare brackets. For a number of problems, I have included aNotethat adds an explanatoryextension of the solution or relates the solution to material in the text. These notes are alwayspreceded by the wordNotein bold type. Most of the problems in the text relate directlyto the material covered. They are meant to extend and deepen the students knowledge ofEM, and to achieve utility with the necessary mathematics. The notes I have added to someproblems are meant to emphasize the connection with the text.The solutions in this manual represent my teaching philosophy in doing the problems.You may have a different approach for some problems. In that case, just use my solutionsas one way of doing things.I have tried to keep virtuoso problems, or problems that areused to learn new things, out of the text. I expect that professors will add some of their ownproblems, in line with their specialties or geared to their student’s interests. I don’t havesolutions for those problems.This is an “Instructors Solutions Manual”. As much as possible, I do not think it wouldbe helpful to students to make it available to them in toto.My feeling is that strugglingthrough the problems, at least once, is an effective tool for absorbing the course material,and also the math techniques needed throughout physics. I have tried to keep the problemsin the text at a level that will challenge, but not discourage, the students. In the text, I tryto lead the students through the material, but the problems are meant to help them see howto work things out for themselves.I hope that the text, along with this instructor’s solution manual, will make the teachingof EM more pleasureable and rewarding for you and your students.Jerrold Franklin

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Contents1Foundations of Electrostatics12Further Development of Electrostatics143Methods of Solution in Electrostatics214Spherical and Cylindrical Coordinates365Green’s Functions486Electrostatics in Matter537Magnetostatics638Magnetization and Ferromagnetism81iii

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Chapter 1Foundations of ElectrostaticsProblem 1.1:(a)The force on the upper right hand charge due to the other three charges at the cornersof a square with sides of lengthLisF=q2L2ˆi+q2L2ˆj+q22L2(ˆi+ˆj√2)=q2[1 + 2√22√2L2](ˆi+ˆj).(1.1)The magnitude of this force on any of the four charges isF=q2(1 + 2√2)/2L2.(1.2)1

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2CHAPTER 1.FOUNDATIONS OF ELECTROSTATICS(b)If the four charges are released from rest, we can write for the acceleration of any of thechargesFm=a=dvdt=dvdL′dL′dt ,(1.3)whereL′is the side length of the square at any time during the motion. The center ofthe square remains fixed, and the distance,r, of a charge from the center is related toL′byL′=√2r, sodL′dt=√2drdt=√2v.Then, a=√2v dvdL′,(1.4)and the squared velocity after a long time is given byV2=∫∞L√2a dL′=[q2(1 + 2√2)√2m] ∫∞LdL′L′2=q2(1 + 2√2)√2mL.(1.5)The velocity is the square root of this.Problem 1.2:(a)The four point chargesqare located at the corners of a square with sides of lengthL.The distance from each charge to a pointzabove the square, on the perpendicularaxis of the square, is√z2+L2/2. The horizontal fields cancel, and the magnitude ofthe vertical field is given byEz=Ecosθ=4qz(z2+L2/2)32.(1.6)

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3(b)For small oscillations,z << L, and we can approximate the force on a charge−qasF=−8√2q2zL3=−kz.(1.7)This is a restoring force proportional to the distance with an effective spring constantk= 8√2q2L3,(1.8)which leads to simple harmonic motion with periodT= 2π√mk−= 2π[mL38√2q2]12.(1.9)Problem 1.3:(a)By symmetry, the electric field of a long straight wire is perpendicular to the wire. Itsmagnitude a distance r from the wire is given byEz=∫∞−∞rλdz(z2+r2)3/2=λr∫π/2−π/2cosθdθ= 2λr .(1.10)We made the substitutionz=rtanθin doing the integral. For the configuration oftwo wires a distanceaapart, the electric field isE= 2λ(r−a/2)|r−a/2)|2+ 2λ(r+a/2)|r+a/2)|2.(1.11)(b)The electric field in Cartesian coordinates isEx=2λ(x−a/2)(x−a/2)2+y2+2λ(x+a/2)(x+a/2)2+y2(1.12)Ey=2λy(x−a/2)2+y2+2λy(x+a/2)2+y2.(1.13)

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4CHAPTER 1.FOUNDATIONS OF ELECTROSTATICSProblem 1.4:(a)The field on the axis of the uniformly charged wire, a distance z from the center of thewire, is given forz > L/2 byEz=QL∫+L/2−L/2dz′(z−z′)2=QL[1z−L/2−1z+L/2]=Qz2−L2/4.(1.14)(b)For−L/2< z < L/2, the pointzis a distance (L/2−z) from the end of the wire. Thewire can be thought of as two parts.The part of the wire fromz′=z−(L/2−z) = 2z−L/2 toz′= 2Lis symmetric aboutthe pointz. This means that the field due to that portion of the wire will cancel. Theremaining part of the wire has a lengthL′=L−2(L/2−z) = 2zand a chargeQ′=2zQ/L. The midpoint of this part of the wire is atz0= (2z−L/2−L/2)/2 =z−L/2.Thus the electric field from this part of the wire isEz=Q′[(z−z0)2−L′2/4]=2QzL[(z−z+L/2)2−z2]=2QzL[L2/4−z2].(1.15)

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5Problem 1.5:The parallel component of the electric field a distancedfrom a uniformly charged straightwire of lengthLis given by the integralEx=QL∫x2x1dx′cosθr2,(1.16)wherex1andx2are the two endpoints of the wire. Letx′=−dcotθdx′=dcsc2θdθr=dcscθ.(1.17)ThenEx=QLd∫π−θ2θ1cosθ dθ=QLd(sinθ2−sinθ1),(1.18)whereθ1andθ2are the angles shown.For the perpendicular component ofE, the same substitution forz, leads toEy=QLd∫π−θ2θ1sinθ dθ=QLd(cosθ2+ cosθ1).(1.19)

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6CHAPTER 1.FOUNDATIONS OF ELECTROSTATICSProblem 1.6:(a)Every point on the uniformly charged ring is the same distance from a point a distancezalong the axis of the ring, and a line from any point on the ring makes the sameangleθwith the z-axis. Thus the electric field atzisEz=Qcosθr2=Qz(z2+R2)32.(1.20)(b)The disk has a surface charge densityσ=Q/πR2. It can be considered as a collectionof rings, each of radiusr′with a chargedq= 2πr′σdr′= 2Qr′dr′R2.(1.21)The electric field a distancezalong the axis of the disk is given as (using part a)Ez=∫R02Qr′R2zdr′(z2+r′2)32=2QR2[1−z√z2+R2].(1.22)(c)(i) Forz= 0+(just above the disk),Ez= 2QR2= 2πσ.(1.23)(ii) Forz >> R, we writeEasEz= 2QR21−(1 +R2z2)−12.(1.24)Using the binomial theorem, we getEz=2QR2[1−(1−R22z2+. . .)].'Qz2.(1.25)This limit, equal to the field of a point chargeQ, can be used as a check on the originalresult.

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7Problem 1.7:(a)Each charge q, at the corner of a square with sides of lengthLis a distance√z2+L2/2from a point z along the perpendicular axis of the square. (See the figure in Prob. 1.2.)Thus the potential at the pointzisφ=4q√z2+L2/2.(1.26)(b)The electric field isEz=−∂zφ=4qz(z2+L2/2)32,as in Prob.1.2.(1.27)Problem 1.8:(a)The potential on the axis of the uniformly charged wire, a distance z from the center ofthe wire, is given forz > L/2 byφ=QL∫+L/2−L/2dz′z−z′=QLln(z+L/2z−L/2).(1.28)(b)The electric field isEz=−∂zφ=QL[1z−L/2−1z+L/2]=Qz2−L2/4,as in Prob.1.4(a).(1.29)Problem 1.9:(a)Every point on the uniformly charged ring is the same distance√z2+R2from a pointa distancezalong the axis of the ring, so the potential atzisφ=Q√z2+R2.(1.30)(b)The uniformly charged disk of radius R has a surface charge densityσ=Q/πR2. It canbe considered as a collection of rings, each of radiusr′with a chargedq= 2πr′σdr′= 2Qr′dr′R2.(1.31)The potential a distancezalong the axis of the disk is given as (using part a)φ=2QR2∫R0r′dr′√z2+r′2= 2QR2[√z2+R2−z].(1.32)(c)The electric field of the ring isEz=−∂z[Q√z2+R2]=Qz(z2+R2)32,as in Prob.1.6(a).(1.33)The electric field of the disk isEz=−2QR2∂z[√z2+R2−z]= 2QR2[1−z√z2+R2],as in Prob.1.6(b).(1.34)

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8CHAPTER 1.FOUNDATIONS OF ELECTROSTATICSProblem 1.10:(a)A uniformly charged spherical shell has a surface charge densityσ=Q4πR2.(1.35)The spherical surface can be thought of as composed of rings of radiusr=Rsinθ,whereθis the angle from the z axis. Each ring has a chargedq== 2πrσR dθ= 12Qsinθ dθ.(1.36)The distance from the plane of a ring to the pointdisz=d−Rcosθ.Thus thepotential at the pointdfrom the center of the sphere isφ=∫dq√z2+r2=Q2∫π0sinθ dθ√(d−Rcosθ)2+R2sin2θ=Q2∫π0sinθ dθ√d2+R2−2Rdcosθ=Q2[√d2+R2−2Rdcosθ−Rd]π0=Q2Rd[(d−R)−(d+R)] =Qd ,d≥R.(1.37)Notethat, ford≤R,φ=Q2Rd[(R−d)−(d+R)] =QR,d≤R.(1.38)Fordoutside the sphere, the potential is the same as that of a point charge.Fordinside the sphere, the potential is constant, and hence the electric field is zero.(b)The potential at a distanced≥Rfrom the center of a uniformly charged spherical shellis ∆q/d, where ∆qis the charge on the shell. The potential does not depend on theradiusRof the shell, as long asd≥R. A uniformly charged sphere can be considereda collection of the uniformly charged shells, so that the potential due to the uniformlycharged sphere will beφ=Q/d,whereQis the net charge on the sphere. (Qis thesum of all the ∆qon each spherical shell.)Notethat the charged sphere need not be uniformly charged as long as its chargedistribution is spherically symmetric.(c)Since the potentials in (a) and (b) are the same as the potential of a point charge, theelectric fields are the same as the electric field of a point charge:E=−∇(Qr)=Qˆrr2,r≥R.(1.39)

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9Problem 1.11:(a)The long straight wire has radiusRand uniform charge densityρ. By symmetry, theelectric field is perpendicular to the axis of the wire. We consider a Gaussian cylinderof lengthLand radiusr, concentric with the wire.Then, forr≤R, Gauss’s lawbecomes∮E·dA=4πQ2πrLE=4π2r2ρLE=2πρr,r≤R.(1.40)Forr≥R, Gauss’s law is∮E·dA=4πQ2πrLE=4π2R2ρLE=2πρR2r,r≥R.(1.41)(b)We have to pick some radius for whichφ= 0. We chooseφ(R) = 0, so that the surfaceof the wire is at zero potential. Then, forr≤R,φ(r) =−∫rRE·dr=−∫rR2πρrdr=πρ(R2−r2).(1.42)Forr≥R,φ(r) =−∫rRE·dr=−∫rR2πR2ρdrr= 2πρR2ln(R/r).(1.43)Problem 1.12:(a)By symmetry, the electric field of the uniformly charged hollow sphere is in the radialdirection. We consider a Gaussian sphere of radiusr, concentric with the hollow sphere.Then, forr≤R, Gauss’s law becomes∮E·dA=4πQπr2E=0E=0.(1.44)Forr≥R, Gauss’s law is∮E·dA=4πQπr2E=4πQE=Qr2.(1.45)

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10CHAPTER 1.FOUNDATIONS OF ELECTROSTATICS(b)The electric field forr≥Ris the same as that for a point charge, so the potential isthe same as the potential of a point charge:φ(r) =Qr .(1.46)Forr≤R,E=0, so the interior of a uniformly charged shell is an equipotential withφ=QR .(1.47)Problem 1.13:(a)The charge density inside the uniformly charged sphere isρ=3Q4πR3.(1.48)By symmetry, the electric field is in the radial direction. We consider a Gaussian sphereof radiusr, concentric with the uniformly charged sphere. Then, forr≤R, Gauss’slaw becomes∮E·dA=4πQπr2E=[3Q4πR3] [4πr33]E=QrR3.(1.49)Forr≥R, Gauss’s law is∮E·dA=4πQπr2E=4πQE=Qr2.(1.50)(b)The electric field forr≥Ris the same as that for a point charge, so the potential isthe same as the potential of a point charge:φ(r) =Qr ,r≥R.(1.51)Forr≤R,φ(r)−φ(R)=−∫rRE·drφ(r)=QR−QR3∫Rrrdr=QR[1−12R3(r2−R2)]=Q2R(3R2−r2).(1.52)

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11Problem 1.14:(a)Forφ=qe−μr/r, the electric field isE=−q∇(e−μr/r) =−qe−μr∇(1r)−qr∇(e−μr)=−qe−μr(−ˆrr2−μˆrr)=qˆre−μrr2(1 +μr).(1.53)(b)The charge density is given byρ=14π∇·E=q4π∇·[e−μr(rr3+μrr2)],(1.54)where we have writtenEin a more convenient form for taking its divergence.Wedifferentiate each term in turn, leading toρ=q4π{e−μr[∇·(rr3)+∇·(μrr2)]+[rr3+μrr2]·∇(e−μr)}=qe−μr4π[4πδ(r) + 3μr2−2μr2−μr2−μ2r]=qδ(r)−μ2qe−μr4πr.(1.55)Notethat above we isolated the term∇·(r/r3), because we knew that it equals4πδ(r).This correctly accounted for the singular behavior at the origin.The termqδ(r) corresponds to a point chargeqat the origin.The other part of Eq. (1.55)represents a negative charge distribution surrounding the point charge.(c)Gauss’s law is∮E·dA= 4πQ.(1.56)We choose a sphere of radiusRas our Gaussian surface.The integral ofE·dSoverthe surface of the sphere is∮E·dA= 4πR2Er(R) = 4πqe−μR(1 +μR),(1.57)where we have takenErfrom Eq. (1.53) The charge within the Gaussian sphere isgiven (using Eq. (1.55) byQ=∫r≤Rρdτ=q−4π∫R0μ2qe−μrr2dr4πr= 4πqe−μR(1 +μR),(1.58)which agrees with Eq. (1.57), and Gauss’s law is satisfied. (The last integral above wasdone using integration by parts.)Notethe importance of the proper treatment of the delta function at the origin.

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12CHAPTER 1.FOUNDATIONS OF ELECTROSTATICSProblem 1.15:(a) F= (r×p)(r·p), withpa constant vector.∇×F=∇×[(r×p)(r·p)]=(r·p)[∇×(r×p)]−(r×p)×[∇(r·p)]=(r·p)[(p·∇)r−p(∇·r)]−(r×p)×p=(r·p)(p−3p)−p(r·p)+p2r=p2r−3p(r·p).(1.59)∇·F=∇·[(r×p)(r·p)]=(r·p)[∇·(r×p)] + (r×p)·[∇(r·p)]=(r·p)[p·(∇×r)] + (r×p)·p=0 +r·(p×p) = 0.(1.60)(b) F= (r·p)2r, withpa constant vector.∇×F=∇×[(r·p)2r]=(r·p)2(∇×r)−r×[∇(r·p)2]=0−r×[2(r·p)p]=−2(r·p)(r×p)(1.61)∇·F=∇·[(r·p)2r]=(r·p)2(∇·r) +r·[∇(r·p)2]=3(r·p)2+r·[2(r·p)p]=5(r·p)2.(1.62)

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