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instructor s solutions manual introduction to electrodynamics third edition david j griffiths


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errata instructor s solutions manual introduction to electrodynamics 3rd ed author david griffiths date september 1 2004 · page 4 prob 1.15 b last expression should read y 2z 3x · page 4 prob.1.16 at the beginning insert the following figure · page 8 prob 1.26 last line should read from prob 1.18 × va -6xz x 2z y 3z 2 z · × va x 6xz y 2z 2 z 3z -6z 6z 0 · page 8 prob 1.27 in the determinant for ×f 3rd row 2nd column change y 3 to y 2 · page 8 prob 1.29 line 2 the number in the box should be -12 insert minus sign · page 9 prob 1.31 line 2 change 2x3 to 2z 3 first line of part c insert comma between dx and dz · page 12 probl 1.39 line 5 remove comma after cos · page 13 prob 1.42c last line insert after z · page 14 prob 1.46b change r to a · page 14 prob 1.48 second line of j change the upper limit on the r integral from to r fix the last line to read 4 -e-r r 0 4e-r 4 -e-r e-0 4e-r 4 · page 15 prob 1.49a line 3 in the box change x2 to x3 1


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· page 15 prob 1.49b last integration constant should be lx z not lx y · page 17 prob 1.53 first expression in 4 insert so da r sin dr d · page 17 prob 1.55 solution should read as follows problem 1.55 1 x z 0 dx dz 0 y 0 1 v · dl yz 2 dy 0 v · dl 0 2 x 0 z 2 2y dz -2 dy y 1 0 v · dl yz 2 dy 3y z dz y2 2y2 dy 3y 2 2y2 dy 0 v · dl 2 1 2y 3 4y 2 y 2 dy 2 y4 4y 3 y2 2y 2 3 2 0 1 14 3 3 x y 0 dx dy 0 z 2 0 v · dl 3y z dz z dz 0 v · dl 2 z dz z2 2 0 -2 2 total v · dl 0 14 3 -2 8 3 meanwhile stokes thereom says v · dl ×v · da here da dy dz x so all we need is ×vx y 3y z z yz 2 3 2yz therefore ×v · da 1 0 3 2yz dy dz 2y 1 2 2 2 1 0 32 2y 2y dy 8y 2 10y 6 dy 1 8 3 8 4 -y 3 y 5y 6y 0 -1 3 5 6 8 3 1 sin 2-2y 3 0 2 2yz dz 1 4y 3 0 dy · page 18 prob 1.56 change 3 and 4 to read as follows 3 r sin y 1 so r 2 tan-1 1 2 v · dl dr -1 sin2 cos d 2 0 r cos2 dr r cos sin r d cos3 cos sin sin3 d cos sin therefore 0 cos sin cos d d sin2 sin2 cos cos2 sin2 d 3 d 2 sin sin cos2 sin v · dl /2 1 cos d sin3 2 sin2 0 /2 1 5 1 1 2 2 · 1/5 2 · 1 2 2 2


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4 0 2 r 5 0 v · dl r cos2 dr 4 r dr 5 0 0 5 4 4 r2 v · dl r dr 5 5 2 5 4 5 · -2 5 2 total v · dl 0 3 +2-2 2 3 2 · page 21 probl 1.61e line 2 change z to +z z z · page 25 prob 2.12 last line should read since qtot 4 r3 e 3 1 4 0 q r3 r as in prob 2.8 · page 26 prob 2.15 last expression in first line of ii should be d not d phi · page 28 prob 2.21 at the end insert the following figure vr 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0.5 1 1.5 2 2.5 3 r q 4 0r in the figure r is in units of r and v r is in units of · page 30 prob 2.28 remove right angle sign in the figure · page 42 prob 3.5 subscript on v in last integral should be 3 not 2 · page 45 prob 3.10 after the first box add q2 4 0 1 1 1 x y [cos x sin y 2 2 2 b2 2 2a 2b 2 a sin b a2 b2 x b 1 2 b a2 b2 3/2 y f where cos a a2 b2 f q2 16 0 a 1 2 a a2 b2 3/2 3


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w q2 -q 2 q2 -q 2 1 1 4 4 0 2a 2b 2 a2 b2 32 0 1 1 1 a b a2 b2 · page 45 prob 3.10 in the second box change and to an · page 46 probl 3.13 at the end insert the following comment technically the series solution for is defective since term-by-term differentiation has produced a naively non-convergent sum more sophisticated definitions of convergence permit one to work with series of this form but it is better to sum the series first and then differentiate the second method · page 51 prob 3.18 midpage the reference to eq 3.71 should be 3.72 · page 53 prob 3.21b line 5 a2 should be 1 2r 4 0r next line insert r2 after · page 55 prob 3.23 third displayed equation remove the first · page 58 prob 3.28a second line first integral r3 should read r2 · page 59 prob 3.31c change first v to w · page 64 prob 3.41a lines 2 and 3 remove 0 in the first factor in the expressions for eave in the second expression change to q · page 69 prob 3.47 at the end add the following alternatively start with the separable solution v x y c sin kx d cos kx aeky be-ky note that the configuration is symmetric in x so c 0 and v x 0 0 b -a so combining the constants v x y a cos kx sinh ky but v b y 0 so cos kb 0 which means that kb ±/2 ±3/2 · · · or k 2n 1 2b n with n 1 2 3 negative k does not yield a different solution the sign can be absorbed into a the general linear combination is v x y n=1 an cos n x sinh n y and it remains to fit the final boundary condition v x a v0 n=1 an cos n x sinh n a 4


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use fourier s trick multiplying by cos n x and integrating b b v0 -b cos n x dx n=1 an sinh n a -b cos n x cos n x dx v0 so an 2 sin n b an sinh n a bn n ban sinh n a n n=1 2n 1 2 1n so 2v0 sin n b but sin n b sin b n sinh n a v x y 2v0 b 1n n=1 sinh n y cos n x n sinh n a · page 74 prob 4.4 exponent on r in boxed equation should be 5 not 3 · page 75 prob 4.7 replace the defective solution with the following if the potential is zero at infinity the energy of a point charge q is eq 2.39 w qv r for a physical dipole with -q at r and +q at r+d r+d u qv r d qv r q [v r d v r q r e · dl for an ideal dipole the integral reduces to e · d and u -qe · d -p · e since p qd if you do not or cannot use infinity as the reference point the result still holds as long as you bring the two charges in from the same point r0 or two points at the same potential in that case w q [v r v r0 and u q [v r d v r0 q [v r v r0 q [v r d v r as before · page 75 prob 4.10a 1 r3 should be 1 r2 · page 79 prob 4.19 in the upper right box of the table f for air there is a missing factor of 0 · page 91 problem 5.10b in the first line µ0 i 2 /2 should read µ0 i 2 a/2s in the final boxed equation the first 1 should be a s · page 92 prob 5.15 the signs are all wrong the end of line 1 should read right the middle of the next line should read left in z z the first box it should be n2 n1 and in the second box the minus sign does not belong 5


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· page 114 prob 6.4 last term in second expression for f should be bx z z plus not minus · page 119 prob 6.21a replace with the following the magnetic force on the dipole is given by eq 6.3 to move the dipole in from infinity we must exert an opposite force so the work done is r r u f · dl m · b · dl -m · br m · b i used the gradient theorem eq 1.55 as long as the magnetic field goes to zero at infinity then u -m · b if the magnetic field does not go to zero at infinity one must stipulate that the dipole starts out oriented perpendicular to the field · page 125 prob 7.2b in the box c should be c · page 129 prob 7.18 change first two lines to read b · da b µ0 i µ0 ia 2s 2 s+a s ds µ0 ia ln s 2 s+a s e iloop r dq dq d µ0 a di r ln1 a/s dt dt 2 dt µ0 a µ0 ai ln1 a/s di q ln1 a/s 2r 2r · page 131 prob 7.27 in the second integral r should be s · page 132 prob 7.32c last line in the final two equations insert an i immediately after µ0 · page 140 prob 7.47 in the box the top equation should have a minus sign in front and in the bottom equation the plus sign should be minus · page 141 prob 7.50 final answer r2 should read r2 · page 143 prob 7.55 penultimate displayed equation tp should be · · page 147 prob 8.2 top line penultimate expression change a2 to a4 in c in the first box change 16 to 8 · page 149 prob 8.5c there should be a minus sign in front of 2 in the box · page 149 prob 8.7 almost all the r s here should be s s in line 1 change a r r to s r in the same line change dr to ds in the next line change dr to ds twice and change to in the last line change r r s to s dr to ds and to but leave r as is r s 6


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· page 153 prob 8.11 last line of equations in the numerator of the expression for r change 2.01 to 2.10 · page 175 prob 9.34 penultimate line n3 /n2 not n3 /n3 2 2 2 · page 177 prob 9.38 half-way down remove minus sign in kx ky kz 2 c · page 181 prob 10.8 first line remove ¿ · page 184 prob 10.14 in the first line change 9.98 to 10.42 · page 203 prob 11.14 at beginning of second paragraph remove ¿ · page 222 prob 12.15 end of first sentence change comma to period · page 225 prob 12.23 the figure contains two errors the slopes are for v/c 1/2 not 3/2 and the intervals are incorrect the correct solution is as follows · page 227 prob 12.33 first expression in third line change c2 to c 7


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table of contents chapter 1 chapter 2 chapter 3 chapter 4 chapter 5 chapter 6 chapter 7 chapter 8 chapter 9 chapter 10 chapter 11 chapter 12 vector analysis electrostatics special techniques electrostatic fields in matter magnetostatics magnetostatic fields in matter 1 22 42 73 89 113 125 146 157 179 195 and relativity 219 electrod ynamics conservation electromagnetic laws waves potentials and fields radiation electrodynamics


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chapter 1 vector problem 1.1 analysis a from the diagram ib ci coso3 ibi coso1 ici coso2 multiply by iai iaiib ci coso3 iaiibi coso1 iaiici coso2 so a b c a.b a.c dot product is distributive iaiib ci sin 03 n iaiibi sin 01 n iaiici sin o2n if n is the unit vector pointing out of the page it follows that axb e axb axe cross product is distributive ici sin 82 similarly ib ci sin 03 ibi sin 01 ici sin o2 mulitply by iai n iblsin81 a b for the general case see g e hay s vector and tensor analysis chapter 1 section 7 dot product and section 8 cross product problem 1.2 the triple cross-product is not in general associative for example suppose a and c is perpendicular to a as in the diagram then b xc points out-of-the-page and a xb xc points down and has magnitude abc but axb 0 so ax b xc 0 :f axbxc k-hb bxc iaxbxe z problem 1.3 a 1x 1y h a /3 b 1x 1y hi b /3 coso x y a.b +1 1-1 1 abcoso /3/3coso 10 cos-1t 70.5288° problem 1.4 i the cross-product of any two vectors in the plane will give a vector perpendicular to the plane for example we might pick the base a and the left side b a -1 x 2 y 0 z b -1 x 0 y 3 z 1


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2 chapter 1 vector analysis x axb y z 2 0 1 6x 3y 2z -1 0 3 this has the right direction but the wrong magnitude to make a unit vector out of it simply divide by its length iaxbi=v36+9+4=7 7x 7y 7z axb 3 2 ft iax bi 16 a i -1 i problem 1.5 ax ay az bycz bzcy bzcx bxcz bxcy bycx x[aybxcy bycx azbzcx bxcz yo zo i ll just check the x-component the others go the same way xaybxcy aybycx azbzcx azbxcz yo zoo ba.c ca.b [bxaxcx aycy azcz cxaxbx ayby azbz x 0 y 0 z xaybxcy azbxcz aybycx azbzcx yo zoo they agree problem 1.6 axbxc x y z axbxc bxcxa cxa-xb so axbxc axbxc -bxcxa ba.c ca.b ca.b ac.b ab.c bc.a ab.c o ca.b if this is zero then either a is parallel to c including the case in which they point in oppositedirections or one is zero or else b.c b.a 0 in which case b is perpendicular to a and c including the case b 0 conclusion:axbxc ax b xc either a is parallel to c or b is perpendicular to a and c problem 1.7 4x+6y+8z 2x+8y+7z !2x-2y zl yl4 ;z 4 1 @j 12a 3x 3y 3z 2a lai :rroblem 1.8 a a.yby a.zbz cos cpay sin cpaz cos cpby sin cpbz 2 cos2 cpayby sin cpay cos cpaz sin cpby cos cpbz sincpcoscpaybz azby sin2 cpazbz sin2 cpayby sin cpcoscpaybz azby cos2 cpazbz cos2p sin2 cpayby sin2 cp cos2pazbz c c b ax ay az ~i=laiai .d d th i a2 a2 a2 is equa s x y z provz e i ayby 3 azbz 2 2 3 ~i=l 3 j=lrijaj k=lrikak ~j,k ik 3 irijrikajak lji=l ij ~3 r r i 0 if if j =k j =i-k moreover if r is to preserve lengths for all vectors a then this condition is not only sufficient but also necessary for suppose a 1,0,0 then ~j,k i rijrik ajak ~i rilril and this must equal 1 since we -2 -2 -2 3 3 want ax+ay+az 1 likewise ~i=lri2ri2 ~i=lri3ri3 1 to check the casej =i choose a 1,1,0 k then we want 2 ~j,k i rijrik ajak ~i rilril ~i ri2ri2 ~i rilri2 ~i ri2ril but we already know that the first two sums are both 1 the third and fourth are equal so ~i rilri2 ~i ri2ril 0 and so on for other unequal combinations of j k in matrix notation rr 1 where r is the transpose of r.


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3 problem 1.9 y x z looking down the axis x a 1200 rotation carries the z axis into the y z axis y into x y and x into z x so ax az ay ax az ay 001 r problem i 1 0 0 0 1 0 1.10 a no change ax ax ay ay az az b a -t i -a i in the sense ax -ax ay -ay az -az that is if c axb ie -t c i no minus sign in contrast to c axb -t a x b axb behavior of an ordinary vector as given by b if a and bare pseudo vectors then ax b -t a x b axb so the cross~product of two pseudovectors is again a pseudovector in the cross~product of a vector and a pseudovector one changes sign the other doesn t and therefore the cross-product is itself a vector angular momentum l rxp and torque n rxf are pseudovectors d a bxc -t a «-bx c -a bxc changes sign under inversion of coordinates problem 1.11 avf bv f cv f problem so if a a bxc then a -t i -a a pseudoscalar i 2xx 3y2y 4z3z x 3x2y2z4 2xy3z4 y 4x2y3z3 z exsinylnzx excosylnzy ex sinyl/z z 1.12 a vh 1o 2y 6x 18x 2x 8y 28y vh 0 at summit so 2y 6x 18 0 2x 8y 28 0 6x 24y 84 0 2y 18 24y 84 o 22y 66 y 3 2x 24 28 0 x -2 top is 13 miles north 2 miles west of south hadley b putting in x -2 y 3 h 10 12 12 36 36 84 12 1720 ft c putting in x i 1 y 1 vh 10 2 6 18x 2 8 28 y 10 22 x 22y 220 x y i ivhi 220v 2 j1311 ft/mile i direction northwest


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4 problem chapter 1 vector analysis 1.13 x x x y y y z zl zj vex x 2 y y 2 z zi2 a v 2 tx x_x 2 y_y 2 z_z 2jx+tyoy tzoz 2x_x x+2y_y y+2z_z z 2 b vk tx x x 2 y j y 2 z zi2 x tyo y tz o 1 -j 1 1 -j 2 2 x-x x-2 2 2y-y y-2 2 2 z-z z 2 -o x x x y y y z zlii 1/1-3 i/1-2i z c /x n n1-n-lfi n1-n-lh21-x n1-n-lix so v1-n n1-n-l i.1 i problem 1.14 cosif y z -y +y sin if +z z sinif multiply cos if multiply by sinif ysinif +y by cos if z cos if -y sin if>cos if z sin2 if sin cos z if if cos2 if zsin2if cos2if» z likewise,ycosif zsinif y so cosif sinif ~v sinif cosif therefore vjy u m~v +cosif vjy +sinif vjz so v i transforms add ysinif zcosif v i z !li 0 sm l v oz oy oz oz oz y cos l v z as a vector qed problem 1.15 bv.vb tx x2 ty 3xz2 tz 2xz 2x 0 2x o tx xy ty 2yz tz 3xz y 2x 3x problem v.v tx y2 ty2xy z2 tz 2yz 0 2x 2y 2x y 1.16 tx ty tz 5 3 0-2 tx3 [xx2 +y2 +z2 ty 5 3 [yx2 +y2 +z2 tz 5 [zx2 +y2 +z2 3r-3 x -3/2 22x 3r-5x2 y2 z2 0-2 y -3/20-22y 0-2 z -3/2o-22z 3r-3 3r-3 o this conclusion is surprising because from the diagram this vector field is obviously diverging away from the 0 everywhere except at the origin but at the origin our calculation is no good since r 0 and the expression for v blows up in fact v.v is infinite at that one point and zero elsewhere as we shall see in sect 1.5 problem 1.17 origin how then can v.v o the answer is that v.v vy bl ~oz !ljl !ljl bl !ljl 8z sm l use resuit m p ra.b 114 8y 8y cos l o g sm l oy 8y 8z 8y cos l 8y 8y 8z oy cog sin if» cogif !ljl cos !ljl sin if» sin if 8y 8z 8y oz cosif>vy+sinif>vz vz -sinif>vy +cosif>vz if if 8z oz sin if cosif 8z 8y 8z ~8 8z oz sin if 8y 8z oz 8z ljl !ljl 8 cosif v sin if i cos2 cosif» sin if sin if cosif» cosif so 8y 8z 8y 8y 8y i sin cos if !ljl sin cos !ljl sin2 if sin2 i sin cos i 8z i i i 8z 8z i


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5 oy sinif>cosif cos2 if 8z a a cos2 if sin2 if» sin2 if cos2 if 8z 8y problem 1.18 8y 8z a vxva x 0 x2 8x 8 ox y 0 3xz2 0 8y z 0 oy 2xz 0 oz oz xo 6xz yo 2z z3z2 0 1-6xz x 2z y 3z2z.1 x y z b vxvb xo 0 oz 2yz 2y yo 3z zo x 1-2yx 3zy xz xy 2yz 3xz x c vxvc problem v or v i !ix y2 0 oy 2xy z2 y z x2z 2z yo 0 z2y 2y [qj 1.19 z3 yx xy or v yzx xzy xy zjor v 3x2z x 3y x3 3xz2z sin x cosh 1.20 y x cos x sinh y yj etc problem i vfg 8 x 8 y o z t gu x j gu y t g¥zz j x ~y ~z +gux uy ¥zz jvg gvi qed ivv axb !ix aybz bx ay ~ox bz az 8bv 8x 8x azby !ly azbx axbz !lz axby aybx b y ml a z 8bz b x ml ax 8x 8y 8y oy b z oa oy § 8z ml +a x §oz b y ml a y 8b b x 8z 8z 8z b y ml ml bz 8y oz 8z 8z oz oa a 8y x 8y qed -ay 8tz az 8~v o b vxa a vxb v vx fa o o v» 8 8 » y 8 v 8 »z x jm az j a y8z x j8a a z jmaz y 8y 8y 8z oz oz ox 8x j a y j oaz -a x z 8z 8x 8y 8y ml ml j m 8az z x fu fu y a a jvxa problem 1.21 a a 1 b ay¥z azu x azu az¥zy axu ayu z ax vi qed a x~ob a y~8b a zfu x 8b a x a 8x y a oy a z~8bv ay~obv z z 8z azfu 8bv y a b f r xx+yy+zz let s just do the x component ylx2+y2+z2 f 1f]x x !ix y!ly z tz ylz2;y2+z2


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6 chapter 1 vector analysis {x [j +x 32x +yx 32y +zx 32z x3 xy2 xz2 x2 y2 z2 i o same goes for the other components hence f v f =0 i c va.v vb x2tx 3xz2 2xz tz xyx 2yzy 3xzz ty x2yx oy 3zz 3xz2xx 2zy oz 2xz ox 2yy 3xz x2y 3x2z2 x 6xz3 4xyz y 3x2z 6x2z z problem ii [va.b x [axvxb x i x2 y 3z2 x 2xz 3z2 2y y 3x2zz i 1.22 :x axbx ayby azbz ayvxbz oy o~z bx a °:xz a by ay °:x ~bz az [bx vxa x =b y ax azvxby -b z oz ay°:x ° vz az°fzz ax a.vb]x ax tx ay ty az tjbx ax °:xz ay ° vz az °fzz b.v a]x b x oaz b y oaz b z oaz ax oy oz so [axvxb bxvxa a yox aobz ob y oy a z obz a z !ll oz ax +a x +a y obz !ilk +a zh +b x obz b oaz +a x ax x ax +b z tz a.vb]x b yox oa +b y oaz obz +b oa +a y ax tv tv ax tz b y oaz b oaz b z oyz oz ax +b zh m ob y ax +a z ll tz tzaxby ax tz tv tv [va.b x same for y and z tyaxbz oaz oy vi [vxaxb x by ax bx ay obz ml b x az !ll oy oy oy oz oz oz tyaxby aybx tzazbx axbz oaz oz bz a x !ll oz oz a.vb av.b bv.a x +b y oaz +b z oaz -a x obz -a y obz -a z obz +a x obz ll x ox oy oz ax oy oz ax oy ml b oaz +a +ob !ll +b y oy x tx 7fx oy oz x tjx tjx oy oz a y oy a z oz b z m oz -b oaz -b x 8az ax oy [vxaxb x same for y and z problem 1.23 vu /g txu/gx tyu/gy tzu/gz gu-f g f gu-f a a v a/g g l 1 r lx l lz f £i.x £i.y y 2 x g2 a y g2 z ox oy a oz ax oy a £i.z oz gvf-fvg g2 qed g az g2 1 txax/g tyay/g tzaz/g g a g a g2 ml [g oaz ax oa oy oz g2 a x £i a y £i a z £i ax oy oz g2 q ed



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