Griffiths電動力學習題解答 Chapter 4 Electrostatic F.pdf

Chapter 4 ElectrostaticFields in Matter Problem 4 1 E V x 500 10 3 5x 105 Table 4 1 a 47r 0 0 66x 10 30 so a 47r 8 85x 10 12 0 66x 10 30 7 34X10 41 p aE ed d aE e 7 34x 10 41 5x 105 1 6 x 10 19 2 29X 10 16 m d R 2 29 x 10 16 0 5x 10 10 14 6x 10 6 1To ionize say d R Then R aE e aV ex V Rex a 0 5 x 10 10 1 6 x 10 19 10 3 7 34x 10 41 1108 v 1 Problem 4 2 First find the field at radius r using Gauss law JE da E Qenc or E 4 0 Qenc l r47rqlr 4q a a2 l r Qenc pdT e 2r ar2dr e 2r ar2 ar 07ra3 0a3220 2q a2 a2 rr2 a2 e 2r ar2 ar 2 2 q 1 e 2r a1 2 2a2 Note Qenc r 00 q So the fieldof the electron cloud is Ee 4 0 1 e 2r a 1 2 2 The protonwill be shifted from r 0 to the point d where Ee E the external field 1q 2d a d E 1 e 1 2 2 47r 0d2aa2 Expandingin powers of d a e 2d a 1 2d 2d 2 2d 3 1 2 2 2 3 a2a3 aaa3a 1 1 2 2 r r 1 2 2 dcP dcP d3cP d34 r r 2t 2 2t 4 4 2 4 ad2ad2d3d2d33 a3 4 d 3 3 higher order terms dd2 1 e 2d a1 2 2 aa2 73 74CHAPTER4 ELECTROSTATICFIELDS IN MATTER 1q 4 d3 141 I3I E qd poa 311 oa 471 0 dl 3 a3471 0 3a3371 oa3 Not so different from the uniformsphere model of Ex 4 1 see Eq 4 2 Note thatthis result predicts 4 EO a a3 0 5 X10 10 3 0 09 X 10 30 m3 compared with an experimentalvalue Table 4 1 of 0 66 x 10 30 m3 Ironically the classical formula Eq 4 2 is slightly closer to the empirical value Problem4 3 per Ar Electricfield by Gauss sLaw E da E 471 r2 oQenc EloJ Ar471 r2 dr or E 471 A r4 Ar2 This internal field balances the external field E when nucleus is off center an amount 471 r o44 0 d ad2 4 0 E d V4 oE A So the induced dipole moment is p ed 2ev 0 AVE Evidently Ip is proportionalto El 2 1 For Eq 4 1 to hold in the weak field limit E must be proportional to r for small r which means that p must go to a constant not zero at the origin Ip O 0I nor infinite Problem4 4 rField of q f Induced dipole moment of atom P a E 0QA1I EO r q411 E r2r Field of this dipole at location of q 0 71 inEq 3 103 E 41 13 2aq2 to the right 7I 0r471 or Force on q due to this field IF 2a 4 q 2 13I attractive 7I 0r Problem4 5 Field of PI at P2 0 71 2in Eq 3 103 E1 4 PI39 points down 7I or Fieldof P2 at PI 0 71 in Eq 3 103 E2 4 P2 3 2f points to the right 7I or I 2PIP2 I Torque on PI N1 PIX E2 4 3 pomts mto the page 7I or Problem4 6 a Use image dipole as shown in Fig a Redraw placing Pi at the origin Fig b E P 471 0 2z 3 2cosOf sinO9 P pcosOf psinO9 10 o Pif Z 2 N PX Ei 471 2Z 3 cos0f sin09 x 2cos0f sin09 p2 AA 4r 0 2z 3 cosOsinO4J 2sinOcosO 4J p2 sin0 cos0A 471 0 2z 3 4J out of the page b 75 p2 sin 20 But sin 0cos0 1 2 sm 20 soIN 4m o 16z3 out of the page For0 0 r 2 N tends to rotate p counterclockwise for r 2 0 r N rotates p clockwise Thus the stableorientation is perpendicular to the surface eithertor t Problem 4 7 Say the field is uniform and points in the y direction First slide p in from infinity along the x axis thistakes no work since F is J dl If E is not uniform slide p in along a trajectoryJ the field Now rotate counterclockwise into final position The torque exerted by E is N pxE pEsinOz The torque we exert is N pEsinO xclockwise and dOis counterclockwise so the net work done by us is negative U J 2 pE sin OdO pE cosO 1 2 pE cosO cos pE cos0 p E Qed Problem4 8 U pI E2 but E2 r 3 p2 f f P2 SOU r PI P2 3 pI f p2 f Qed Problem4 9 1q qxx yy zz a F p V E Eq 4 5 E 4 r 4 222 3 2 rEO r rEOx y z y tE O P p 888 qx Fx Px P pz 8xY8y8z4 rEO X2 y2 Z2 3 2 q 132x 32y 4 rEO Px x2 y2 Z2 3 2 2x X2 y2 Z2 5 2 py 2x X2 y2 Z2 5 2 32z q Px3x q p3r p r pz 2x X2 y2 Z2 5 2 4 rEO r3 s Pxx Pyy pzz 4 rEOr3 r5x F I 4 1 p 3 p f f rEo r b E 41 3 p f f p 4 113 3 p f f p This is from Eq 3 104 the minus signs rEO r rEO r arebecause r points toward p in this problem F qE 1 41 q3 3 p f f p rEO r Notethat the forces are equal and opposite as you would expect from Newton s third law Problem 4 10 1 8212 a Ub P n kR Pb V p 3 8 r kr 3kr rrr b For r R same as if all chargeat center but Qtot kR 4 rR2 3k t rR3 0 so IE 0 1 76CHAPTER4 ELECTROSTATICFIELDS IN MATTER Problem4 11 Pb 0 ab P il P plus sign at one end theone P points toward minus sign at the other theone P points away from i L a Then the ends look like point charges and the whole thing is like a physical dipole Qflength Land charge P rra2 See Fig a ii L a Then it s like a circular parallel plate capacitor Field is nearly uniform inside nonuniform fringing field at the edges See Fig b iii L a See Fig c ppp a Like a dipole b Like a parallel platecapacitor c Problem4 12 v 4 EOJI jdT p 4 EOJ dT But the term in curly brackets is precisely the field of a uniformly charged sphere divided by p The integral was done explicitly in Prob 2 7 and 2 8 I R3A I R3P cos BI R I 4 3 R3Pf r R 3 or2P r 3 or2 r 1 t 1411 0rSo V r B dT 4W O p 4 3 wR p r r R I P r I owoo 1 r R 411 0R33 0 Problem4 13 Think of it as two cylinders of opposite uniform charge density p Inside the field at a distance s from the axis of a uniformly charge cylinder is given by Gauss s law E211 se OP1l S2e E p 2 0 s For two such cylinders one plus and one minus the net field inside is E E E p 2fO s s But s s d so E l pd 2 0 Iwhere d is the vector fromthe negative axis to positiveaxis In this case the total dipole moment of a chunk of length e is P 1I a2e p7ra2e d So pd P and IE P 2 0 Ifor s a 77 Outside Gauss s law gives E27r8 1 p7ra2 E 1 2a2 for one cylinder For the combination E 0 0 s E E 1 2 a2 t iL where a Problem 4 14 Total charge on the dielectric is Qtot is O b da Iv Pbdr is P da Iv V p dr But the divergence theoremsays is p da IvV p dr so Qenc O qed Problem4 15 a Pb v p r2 O b P ii P k b atr b r2 orrr2 P r k a at r a Gauss slaw E 4 0Q cr For r b Qenc 0 Prob 4 14 soIE 0 1 Fora r b Qenc k 47ra2 I 47rf2dr 47rka 47rk r a 47rkr soIE k for r 1 b fD da Qfenc O D 0 everywhere D foE P O E l fo P so IE 0 for r b jIE k fOr r for a r 1 2fO totalsurfacechargeabove a a 2 a 2 a 3 2a 3 E 2a total surface charge below a 3 a 2a 3 2 3fO u u 2 u 2 u 3 u 3 u Problem4 19 With no dielectric Co Afo d Eq 2 54 In configuration a with a on upper plate a on lower D a between the plates E a fO in air and E a f in dielectric So V 7 2 1 2 ICa 2frI Ca V d1 1 r Co 1 fr In configuration b with potential difference V E V d so a foE fOV d in air 79 P EOXeE EOXe V d in dielectric so O b EOXeV d at top surface of dielectric O tot EoV d O f O b O f EoXeV d so O f EoV l Xe d EOErV d on top plate above dielectric Q1 AA A VV AEO 1 Er ICb1 ErI Cb V V0 2 O f2 2VEod EOdEr d Co Which isgreater l r l r 2 4 r 1 2 r 4 4 r 1 0 So C C CoCo2l r2 1 r 2 1 r 2 1 D41fr2 p 1fr3 D lpr E pr 3E r for r D pR3 3r2 E pR3 3Eor2 r for r R fOpR3 1 1 RPfopR2P R2 I pR2 1 V 00E dl 3EO 00 3E R rdr 3EO 3E 2 3EO1 2Er Problem 4 21 Let Q be the charge on a length of the inner conductor V QQQ D21fs Q D 2 0 E 2 0 a s b E 2 0 b r c 1fS 1fEOS 1fES f a I b Q dS l C Q dSQ b EO C E dl In In C a21fEO Sb21fdS 21fEO aEb QI21fEOI V In b a l Er In cjb fD da C Problem 4 22 Same method as Ex 4 7 solve Laplace s equation for V n s tj s a subject to the boundary conditions x i V n Voutat s a ii E8 n EO 8 utat s a Hi Vout Eos cos tj for S a Eot y FromProb 3 23 invoking boundary condition Hi 00 V n s tj 2 k ak cosktj bksinktj k l 00 Vout s tj Eoscostj Ls k Ckcosk dk sink k l a air 2 r 00 Y r 1 d x rH d X r l d a dielectric 2v x r l d X rH d r l dX r l d b air Yx 00 7 left ddx b dielectric Yx Er7x Er 1 7X Er 1 7Er7 right d 80CHAPTER 4 ELECTROSTATIC FIELDS IN MATTER 1 eliminated the constant terms by setting V 0 on the y z plane Condition i says I ak akcosk J bk sin k J Eoscos J La k ckcosk J dk sink J while ii says f rLkak l akcosk J bk sink J Eo cos dk sink J Evidentlybk dk 0 for all k ak Ck 0 unless k 1 whereasfork 1 aal Eoa a ICI f ral Eo a 2CI Solving for aI Eo al 1 Xe 2 Eoscos J so n S J 1 Xe 2 Eo x 1 Xe 2 and hence Ein s J x 2 1As in the spherical case Ex 4 7 the field inside is uniform Problem4 23 1Xef OX 1X Po f oXeEo EI 3 Po 3Eo PI f oXeEI 3 Eo E2 PI Eo 9 Evidently En e nEo so E Eo EI E2 er Eo The geometric series can be summed explicitly 001 n x l x n O so 1Eo E 1 Xe 3 which agrees with Eq 4 49 Curiously this method formally requires that Xe b a r b r 0 41T 0 X2 y2 z d 2 x2 y2 z d 2 q r Meanwhile since qt 7 1 r r rrr 1 2q r I for z 0 V r 41T 0 X2 y2 zd 2 I I I I Problem4 26 From Ex 4 5 D O Q r a 0 Q r E 41Ttt r 41T or r a a r b 1 I 1Q2 I l b1 11 1 001 Q2 I 1 I b1 1 1 00 w D EdT 41T r2dr dr 22 41T 2 a r2 r2 obr281T ra orb Q2 1 11 1 Q2 1Xe 81T 0 1 Xe b b 81T 0 1 Xe b Problem4 27 83 Using Eq 4 55 W fJE2 dr From Ex 4 2 and Eq 3 103 E WrR Wtot I 3102 P z 3 R 2cosBf sinBO for fa 2 11 R3 211 P2R3 23100327fa fa R3P 2 J 1 2 2 2 6 4cos B sm B r smBdrdBd jJ 23100r R3p 2 l 1r 1 00111 R3p 2 8211 1 3COS2 B sin BdB4 dr 1 faaRr9100 11 R3P 2 411 R3p2 91003R327100 211 R3 p2 9100 COSB COS3B I 3 3 1 r R so Thisis the correct electrostaticenergy of the configuration but it is not the total work necessary to assemble the system because it leaves out the mechanical energy involved i polarizing the molecules Using Eq 4 58 W JD E dr For r R D foE so this contribution is the same as before Forr V 1L In bJa 41rfOS41rfO 10 j lOr OilPart D E 2 V 2 In bJa fa10fa 41rs41rfS41rf Q h f h fr h h f fr l h f Xeh f where f is the total height Q Xeh f Xeh f C V 2 ln bJa 411 100 211 100 In bJa Thtd l bE4 64 F 1V2 dC 1V2 21rfOXe I V 2 e neupwarLOrCeIS gIvenyq 2 dh 2 In bfa h fOXe Thegravitational force down is F mg p11 b2 a2 gh p b2 a2 g In bJa 84CHAPTER4 ELECTROSTATICFIELDS IN MATTER Problem4 29 8 a Eq 4 5 F2 P2 V EI P2 Ed uy PIAPIA Eq 3 1O3 EI 43 z Therefore 4m or1rfoY y jr PIP2 d 1 A3PIP2A I 3PIP2A I F2 4 d 3 z 4 4 Z or F2 4 z upward 1rfOyY1rfoY1rfor z y To calculateF I put P2 at the origin pointingin the z direction then PI is at r z and it points in the ydirection So FI PI V E2 PI 8 2 1 we need E2 as a function of x y and z yx y O z r I I EE ll 3 P2 r r h AAAA d h Fromq 3 104 2 3 P werer xx yy ZZ P2 P2Y anence 41rfOrr P2 r P2Y E2 3Y XX yy zz x2 y2 Z2 y 3XYX x2 2y2 z2 y 3YZZ 41rfO x2 y2 z2 5 241rfO x2 y2 z2 5 2 2Y 3XYX x2 2y2 Z2 y 3yzz 3xx 4yy 3ZZ 2 3z z FI PI 3r Z 3PIP2 Z 41rfOr541rfO r541rfor4 8E2 8y 8E2 1 8y 0 0 These results areconsistent with Newton s third law FI F2 b From page 165 N2 P2 x EI r x F2 The first term wascalculated in Frob 4 5 the secondwe get from a using r ry P2 X EI PIP2 A 41rfor3 X j F 3PIP2 3PIP2A I N 2PIP2A r x2 ryx 4 4 Z 4 3 x so2 4 3 X 1rfor1rfor1rfor This is equal and opposite to the torque on PI due to P2 with respect to the center of PI see Frob 4 5 Problem4 30 Net force isIto the rightI see diagram Note that the field lines must bulge to the right as shown because E is perpendicularto the surface of each conductor E 85 Problem 4 31 P kr k xx yy zz Pb V p k l 1 1 1 3k 1 Total volume bound charge IQvol 3ka3 1 Jb P il At top surface il z z a 2j so O b ka 2 Clearly IO b ka 21 on all six surfaces T tal surfaceboundcharge IQsurf 6 ka 2 a2 3ka3 1Totalboundchargeis zero if Problem 4 32 f q 1qfqXef D da Qfonc D 4 2 rjE D 4 1 2 P fOXeE 4 1 2 7rr107rfO Xe r7r Xe r Pb V p 47r Xe V q1 eXe 83 r Eq 1 99 j O b P f 47r 1 X e R2 Qsurf Jb 47r R2 IqXe 1 The compensating negative charge is at the center 1 Xe j PbdT l qXe j 83 r dT q 1Xe Xe Xe Problem 4 33 Ell is continuous Eq 4 29 Dl is continuous Eq 4 26 with O f 0 So EXl EX2 DYl DY2 E1EYI f2EY2 and hence tan02 EX2 EY2 EYl E2 Qed tan 01EXl EYlEY2El If1isair and 2 is dielectric tan O2 tan 01 E2 EO 1 and the field lines bend away from the normal This is theopposite of light rays so a convex lens would defocus the field lines Problem 4 34 In view of Eq 4 39 the net dipolemomentat the centeris pi P 1 e p I Xop tp We wantthe potentialproduced by pi at the center and O b at R Use separation of variables 1 00B Outside V r O Lrl lPz cosO Eq 3 72 1 0 1pcosO00 lnstde V r O 4 LAlrIPI cosO Eqs 3 66 3 102 7rEO Err1 0 R L AIRI orBI R2 HAI l 1 V continuousat R B1 1P p 3 R2 47rEoErR2 AIR orB1 47rfOf AIR 8V I av i8r R ar R BI12pcosO I11 L l 1 Rl 2Pz cosO 4R3 LlAIR PI cosO O b 7rEOErEO 1 1 av I I2pcosO I1 p r EoXeE r Xe a Xe 4 R3 lAIR PI cosO EOEOrR 7rEO lOr 86CHAPTER 4 ELECTROSTATIC FIELDS IN MATTER I 1 R 2 lAIRI I XelAIRI I l 1 or 2l l AIRI I XelAIRI I Al 0 1 BI12p 12P PAIR31XePAIR3 Forl l 2 AI Xe A1 BI Xe R3411 fO frR3411 fO frR3411 fOfr2411 fo fr2 AIR3 AIR3 XeP XeAIR3 AIR3 3 Xe XeP 411 fOfr411 fOfr2411 fOfr22411 fOfr Al 2XeP 2 fr l p BI 1 2 fr 1 411 fo R3fr 3 Xe 411 fo Rafr fr 2 411 fOfr fr 2 411 fOfrfr 2 V r O pCOSO r R 411 for2fr 2 Meanwhile for r R V r 0 PcosO 1 prcos 2 fr 1 411 fo frr2411 fOR3freEr 2 P cos fr 1 r3 I421 2 2R3 r R 1I forfrfr Problem4 35 Giventwo solutions VI and EI VVI DI fEd and V2 E2 VV2 D2 fE2 defineV3 V2 VI E3 E2 EI Da D2 DI Iv V VaD3 dr Is V3Da da 0 Va on S so I VV3 D3dr I V3 V D3 dr 0 ButV D3 V D2 V DI PI PI 0 and VV3 VV2 VVI E2 EI E3 so IE3 D3 dr O But D3 D2 DI fE2 EEl fE3 soIf E3 2 dr 0 But f 0 so E3 0 so V2 VI constant But at surface V2 VI so V2 VI everywhere qed Problem4 36 a Proposed potential IVCr Vo 1 If so then IE VV VO f Iin which caseP fOXeVo f in the region z 0 of course Then CTb fOXeVo f ft 1 fO Vo 1 Note ft pointsout of dielectric ft f This CTb is on the surface at r R The flat surface z 0 carries no bound charge since ft z 1 f Nor is there any volume bound charge Eq 4 39 If V is to have the required spherical symmetry the net charge must be uniform CTtot411 R2 Qtot 411 fORVo since Vo Qtot 411 foR so CTtot foVo R Therefore fOVo R on northern hemisphere CT I fOVol R l Xe on southern hemisphere b By construction CTtot CTb CTI foVo R is uniform on the northern hemisphere CTb 0 CTI foVo R on the southernhemisphereCTb foXeVo R so CTI fVo R The potential of a uniformly charged sphere is Vo Qtot CTtot 411 R2 fOVo R2 VoR 411 for411 fOrRforr c Since everything is consistent and the boundary conditions V Vo at r R V 0 at 00 are met Prob 4 35 guaranteesthat this is the solution 87 d Figure b works the same way but Fig a does not on the flat surface P is not perpendicularto ft sowe d get bound charge on this surface spoiling the symmetry Problem4 37 Eext 8 Since the sphereis tiny this is essentiallyconstant and hence P oXe Eext Ex 4 7 27r 081 Xe 3 2 F oXe 8dr oXe 18dr J1 Xe 327r 08d827r 08 1 Xe 32no882J Xe 2 7rR38 2R3 8 1 Xe 347r2 0833 3 Xe7r 083 Problem4 38 The density of atoms is N 4 3 7rR3 The macroscopic field E is Eself Eelse where Eself is the average fieldover the sphere due to the atom itself p o Eelse P No Eelse Actually it is the field at the center not the average over the sphere that belongs here but the two are in factequal as we found in Prob 3 41d Now 1p Eself 47r 0R3 Eq 3 105 so 10 0 NO E 47r 0 R3 Eelse Eelse 1 47r oR3Eelse 1 3 0Eelse So P No 1 N o 3 0 E oXeE andhence No o X