大學(xué)物理第五版課后答案(馬文蔚)

1、 ttPs r r sv v rt tr sv v t tdrdtdrv dtrsvty 22v tdt vtadr vdtrdsdvdtdta aana a aa h,tdldl dtlx l hdxdt2 ltv dtl h222vvlvvx00cos0l h /l22 xx x xxt00ttps x xt ttx x tdx d xpp12122s 和 dt dt2x x x m解 x400t 2s由tpx x x mx x x 40m,則1202 42s x x mx12v 48ms1 ,dtmd x2,a.ts4.0s2t2 t4.0s tttttxttttt解 v va AB20

2、ms2 a 0ABt tBCBAv vDa CD10ms2 Ct tDCtx xv t t1220 xtvms1 xt ytsds (dx) (dy)s ds2 21y 2 x2解 4tt20tt212020r x) y) m22r r r r x y x y m2020s122s Qds 42P0 解 dxdtytv x 10t,v 15tyt v , v v v v ms22100 x0yvxovv320y0 xdvdva x60ms2 ,a 40ms2yxdtdtya a a ms222xyaxaa23tan yx y yy y1122 11y v t aty hv t gt222210

3、20y y 12112hv t at hv t gtt s2222ga001d h y v t gt 0.716m2220 g12h0 h (gatt s22ga12則20解 如2因T 為T,Tyy yy Rcos tR,TT0t)itjTTdr1T22a 2222TTTT2解 ,ds312s1st arctan h 436060s dvdtdxdta和vv t dx vdt 和 vat解 vdvtadtv001v t t v得330 xdx tvt由得001x t t v tx241200 v x001x t t 0.752412 dvvtdta(v)解 ydva AvdtdvdtAvdvA

4、vtdv dtv e )BtAvBv00AvBdy Av e )Bt dt BAAA ydy te ty t (e BtBtBB B200 a axy1x x v t a t2和20 x12 20y解 t v 00vttiv t jaij000drr0rttrij2200 x2 y2 yxm23 tan dv a和tvtvtv2a Rnt 解 vv v 21)而vvvv2a ta 0.90031,RRRRv2R xy,rttvdta a na e aa t tt1v2a nx, y2ty 2t t 12r221dydt22j 2.0i tjat) i2xy221 i jtdvda e ( v

5、v )e 3.58ms ea a a e ms e22y222t2dtdtt t1stxttnnn tv2v v v 4.47 ms則 11.17 m221axy nv v求xyg解 xvt, ygt2 2yx vmgy arctan 12.5oxvgt arctan arctanyvvxgta gsin gsin arctan1.88ms2vtgta gcos gcos arctan9.62 ms2vn xyvv,00yxv x1gr220OP1222200y t x )221gt2r2t0sin22y v x0t Ov解 1x vt ty gt h2222x 2gr x R R 1 222

6、g 0v 0v sin2R 0g xy值解 Oxy1x, y vtsin gt22y tan2x v 2 sstaav a eaetn2n ns-st0解 dsv v btdt0d sv2 (v bt)22a t b a ,0tRR2na b (v bt)24a a a 2n2t0Ra (v bt) 2arctan arctan n0aRbR b (v bt) bt1224R0vt 0bv /b0v2s s s 0bt0sv2n 02R 4bR kt2解 2v2tRt22212dtt 2aa a e e a R R222nttn2 2dt 223000 解 2t3da r 2.30 ms r2

7、2dt121 4222得33 24rt r t4 t222nv v 12 v 122221v1tan75o vh2v 22112211l1hhh2 v uvu解 uarcsindd1.0510 s3dl utu 5.0102 m解 OxyOOxtx- vtvt- vty2d y22y F NFN FNFNm )2FRN F mgaABABAB解 O l11lt 2則cos 22gcos sin cosdt sin sincos cos cossin 0t 解 Oy軸a F - m gm a21222122a F N3 -F -m (ga) N3N2N22a F N3Nm g-F m aAAB

8、F AB F f2 F -F a12a a aa a 2- as2 WF s)F l) 122222s F F 始NNOxy2NnF cos N RhRh FNvv 00v F v102解 v2F sin mF 0RNN0v v02F Fsin 0N1N1v2R1v 0v2F cosFsin 0RN2N2v2R2 ,相當(dāng)于如圖(b)v v v121,而v2 F F N解 v2F 0F mRN1N22R hN22m 4 R v2222F N2222222222F F F m g 2N12N222為F22N2F222N1 t,a t, 0 t a t, t dvdt ta得 vdvtadtv5t2

9、v00dx vxdx tvdt得dtx001x 2t t33tv x m55t t t t2vdvta dt2v得v5s0 vtxdx t得xt tt2 3x55stv 和 x m77 ,vx解 dvt 40mtt v 00得 vdv tt4.0 tvt+ t2 v00 xt x 00 xdx t6.0tt dt2x00 xt t23 F解 dvt F ma m tvdvt dtv v t2得dtm2m0v00v xdx tv 0t dt2又2mx00s xx v t t 467m36m00 FF解 v 20PF FmaPFbv avtv v2-bv mv v)2v 20y 0mdv t dy

10、 vv v e 2ghe/mby/mbv00v0b/mm vv 0m vy ln 5.76mb v0 解 OO一FFm dF F r F rdr rdr2lTTTrF 2 222l2mn2tl F rdrF r l r l r22ll TTF rrTrF0 N5 aFanFmgNFF nnNvdtF F cos mRnNr由 tv v得v90o0 vF mrrN NNNnFNF FFFNNN解 vtRNnftF Nvdvdt2 Rt vv 0R dvvtv0 v20v00v v00Rtv0vR s tvdt tdts ln20Rv t000 v解 F kvrdvmg vmdtdvm v tdt

11、 mvvt ln16.11s0mgvkmg0v0dvdtdv vdydvdymgvvvdvmmg v ydy 0y ln1v 183m故0mg vk kmg00v0vv2 t和y 00g2g PF Prr解 ydvvdvdymgv m m2dtvv1 g v 2ydy 0y lngv22kg v2000 v1 gv 2h y ln02kgmaxvdvmg v m2dyvdv v 1/2 v v 12ydy 0則g v2g00v0 rkv F2rF解 xF rdvF v2 mdtat2m v dvdt2F1 mv2mm v 1v 122tt v1 dvt mln32F則mFv200mv dvm

12、dtxm v 1v24v2 12xdx v1 vx ln 0.144則2mmmFv2F 3F200m F F 為12F 3dxtt用vxv解 xdv2dtN1F k v mg 0N22dvdvx2t12vdvvdxmg k k v012m2x 12mg12F v kv 22N00 xk v k 2x ln201k122k v 51k02x F 0F -Fma 0 2 2 解 F 1maF maarm gmaF ma11T122T2F FT2T2m mF F ga121m mm mT2T21212m m g2m aa a a a a a 122112m mm m1r2r1212a a a 和aa

13、 mgmgm a21rr12211 a a a ABABa a a BABAA mgaA. B ABP AF AP B F F A的2OxOya Aa B ABaABy a a a a BABBAABxABAmgsin cos2a B222B 22222BBxBy2y aacos F N1AB BxBymsin 2 AOxB mgsinma cos maABAF sin 0N1AF a 0N1AF FN1N1mgsincos m gsina a BAsin sin A22a a aBBAA 2mm sin mgcos222a gsinF N1a msin sin B2A2 3 C 與A或D與B

14、 解 x Ft v0 t /v,v2F2.5510 NF F2.55 N55l 5 t v sint 0g1 v sinv sint t 2t 00gg121t2 t t v t t 2v I1FjjIFjj10220t1 O mv v y v sinjI mv jv j 2v sin jI1jjAy002By00 I t2dtt1v2 解 I 230t dt t t 68Ns2 200 I t t t 2 I mv- mvI v21v 240ms1t s I v1m1 mv 21 F P t v v21 vt 2ghtF mg mg1.14103 N mgF 2h/g mg1.1410 N3

15、t F t2tx x t1解 F kAkAI t2Fdt t2dt /2costdt v 即t1t10 t t v v BAI 解 t v v ) t v v )BABAI I Svt v vFtBAFF 2Sv 2.510 N23 解 xgv x1t0 x1012211y t t 111h gt22t12112211h gt2g2111x x v t21122y x m22 搬ABAv m m m v v m AABAAm m v v m BBABBv v B BBABA ABBAB 0v v u mmu人的水平速率的增量為v vv cos umm000t 00g 由于桌面所受的壓力難以直接

16、求出,因此,可轉(zhuǎn)化為求其反作用力,即桌面t 解 Oy,t FNF mlNmt y lP F mlF F而mmm2lllNN t mt t P mgt it m t dmumg maa dt0m t 時(shí)0mm 0解 dmdtumg mam 50a 0dm m g a3.6810 kgs0031dtu mtvtdm umg mvdv umtgdtmv000mv v ulngtm00t dmdt1m m t m605mt 0得6dm/dtmm5mulngt uln 2.4710 ms031mm 6dm/dt00 F x FF F xLx0L00FF F 0 x, x 0 x L 解 L0FF LW

17、LF x dx 00L2001F LW v20 x L v02mFdvdvdxF xm v0Ldt0 W F ds解 W F dx cosdx x222W F x F ) kv 2x ct 3解 x ct 3dxv 222 42/3 4/327llo2/3 4/32/3 7/3700P mg lW F dy 10 mgagydy882J00WPW F sTT WE E 0.53JvKPmmkk 2.49 Nll113W E E mv mv mv2228kk0 F W F ofv203mv20 B 11222112y y NOF ky F ky P mg112211F F P121F P F12

18、ANB P F 2222F P P m m g1212解 x W mgl )W E EE )PE Em) l ) ,PW m) l ) m 1 3 解 k x 0 xdx x xdx m000 x0 解 m mv21m mGmE v G E則2E3R3R26R2KEEE m mE GE3RPE m mm mm mE E E GG GEEE6R3R6RKPEEE 解 1mgR v cos22v2cosF RNF N2 arccos 48.2o32v 3v P C 解 C F Nv2mgcrA11 k l mg r v22 c227mgrk 366 Nm1 l2 解 x v 01 v m v111

19、1 v m v kx22202221x v 0k m 解 vvm 2F 2g hl h112gl 222hvm 5 E /EE /E v /v2 2HeHeHeE /E 即He解 E E He1v2Ev 221HHHEm v v2ee2e有v v eHe111v v 2e22222He 22Hee 1mcosA22ABm2BA12222ABA2271BAA2222 oAAAAP P 12F 01v 2h m gcossin11 m v m gh m v222221m2v 2v cos 2gh cot1m0 解 v v 0mm121v v mgR222vmv v m2v mmm 2v mmm v

20、 v v v v 2gRmmmmm2F mgmRN2mF mg 3N 解 gh h4 ShKdh 110gh 18.88m2 SKv 20 m v m v01 h h2 -4d Sv m v m gh21202h1h m2 mv 21v m 2 01h-4 S 35mh dh gh 23230d0i icmdtiii mx 0解 iimv mv mv 02mv m v mv 02v v v v sin,22m2121mm33v 2.8ms i 2.0ms j則113 解 t m2m m12m1m m12dvxdtx112y212t v F tFdt vm m dv , v tx1m m112x

21、x0012F tFdt 1vm m dv , v ty2m m12yy0012t x y F xdx tdtc1m mcc0012Fx x t 1.5t2212 m mcc012 F ydy tdt及c2m mcc0012Fy y t 1.9t2222 m mcc012 P P F F dt t i t jt120 L d d J 00mvD J O J JJ 00 r r r 解 n d dt 2 nn 13.1rads200tt 1 t t t nn202200 s nnN t 390022 s 解 t s 1e 8.6st/100 d e rads0t/t/22dt t s 6dt 6

22、1e dt 36.9radt/000t N /2 5.87圈 解 22AAH22HJ J 2m d則2H112mHJ /J tan 則2BBJJo 1d1d221222212121122 r mrm J r dmRr2rdr22R20R/22m15r dr mRR32R232R/21J mR解2 2 211 mRR1 2 2J 2 mR 22 R2232 215J J J mR232012 15mR39R mR 2J mR m 22232R22320 解1 F F R TmgF maTa 1h at22 gt2J mR 122h解2 11mgh v 02222v Rv222J mR 20t2J

23、 t 0J nn 10.8sMM0解2 tdt J 00 t J 則0MM0解 122T12T2T2F F a 2m 2m121222212 12T121T11T1T22T22 F RF r J J T1T212,T1T1T2T21gRgr121221212a 212221212212122211212212112221212解 F P F 和1F mgsinmgcos maT1111 1 F P 2m gF m a2T22 2a a r12T2T1,T1T1T2T2a a 211J12m m r1222121212mm g 1sin cos m gJ /r2122212F F 12N 1F

24、F而NNl lM F F d d12lfN100ttt2，nmdlF 3.1410 N 2112 r R Mr F ，2M dt J解 r 2M rF 2r r/R k2fk 2r mg2 R2M dM dr mgRR230 J mR t 2J Rt M 4g 解 MCMJddtC JC t dtJ00C J eCt/ J0 00Jt ln2C d t etCt/ J000 即0Ct 0解 z z 22222t0 2ml 1e sin 2ml sin e2t222t00t 220 etMJ0.解 0v2 R22h02g2g L L 01L RmR2L 1L m R 22 解 J J J 212

25、 J m l/22Jv/l222J ml /122 11J 6m v29.1s122J J1m m212 解 t 有 F t J r1110F t J r222r r 1122J r2 10 2J r J r1221 220 1 01 .01解 v R010 J J 000101J J mR201mR2v 9.5210 s21J mR R020 t 內(nèi)解 s m 10st 0.10kg0 J J mr 2000t s J 0.80J s001J mr021 ur mm解 mur 0mt 2.67sQur0a01b12220J0baa01 r J 即22 0a01c201122202288 o2

26、解 mrv mrv1 12 212Gmm1Gmmv v 22EEr2r1212Gm mr r 12Gm r v 18.1110 msE 2r r r31112r1rv v 6.3110 ms31212 T E J 2E T W EK解 R 61E 2 m R /T 2.1210 J2222292KE2 T2d dTT22 T T22T23E T E T2KKW EKE TM K2n n T 解 J J 010J J 0011J mrJ J r r 0141J 41J000113222221100 MJl212MJ 3Jld ddt d 60od003gs1600l 1E mgl 0.98J2K

27、1E 22K為2E3gs1KJl 解 J J J 1112 2 n nJ 2J 1J 20.0kgm112122n22 11 E J J J 1.3210 J2221422121 解 vv J J J J ll 112301J J lJ l2、2和231231213 J J gl g l g l 202223 4v 2gln v BB1mr 2 22B1mv2 B2解 J J mR 2000B 1211J mgR J mR v202222200BB J 00J mRB20J R22v 2gRB00J mR20 C0v 4C 1J R解 221 J 2 J 2m Rl2和222 2 J J 2m

28、 Rl 1211 2J J 2m Rl 222212l R 114m l v 解 v m01211 2m m v k l l22220 v m m l m m l 0 mk ll22v v 20mm0v l arcsin 0 0m vl 解 cv CF t 21 212x 1441122222232KC1解 F ffC2f1CCR cos RR2F112J mR2C1aFC12R21C10 解 q 12 810 e21maxFq2 1max6eF 4 2g0 解 1 qq1 e2 F e re 3.78N e1 224 r4 r2rr00F e r v1 e22m r 4 r20證 11 e2

29、E v 228 rK0e22303K2解 F 1 qqe21 292200 q P e2r0P E E P P E iLyL證 P E r r201xQ111QE P222000 P E y E dE20r r x2 21xQ1E 44r L222200P 1E lim22l02 r0rL 22解 qS2R sin2 1qE i4x r 2/3220 x，r 1dq1 RcosdE 2R sind244 Rx r 2/332200sincosd20E /2sincosd 24000 P erP cos000 x r 01 2pE 4 x30P 2P 解1 001 2p1 4er cos1 er

30、 cosE 004 x3 4x3x3000解2 E E E2cos2eE 2E cosE 4 r4 x2200r x r 2xrcos22200 xr coscos 0r2exr cos1E 2/304x2x r 2xrcos22000 x r ， 02r 3 2r 2/3x r 2 2/3 x 1x 1 223030 x2x001 r cosE 0 x30 F F E 解 E E P 1 1 E E E i2 x r x00r0i 2 x r x00 F F FEi2 r0 022 r FEi0 0F F P 解 1 2q1q1qE k k k 4 z204 z d4zd2200z kq2

31、111E 4zz 1/1/d zd z22220q2 1 2d d2d d22 1.1. k4zz2zzzz20q 6qd2k4 z40Qqd 21 QE k4 z40S ES sR S SE dS 1q 00 ES ESSS解1 ES ESSSS ER R E22解2 E E e cose sin erS R sind e2r E dS ER sin sin dd22SSER sin d sin d2200R E2 解 E OxyOxy即OABCDEFG 2122CDEOABGF22 E dS Ei E j dSi Ea2AOEF121 E dS E i E j dSi E a1213(RE

32、E1qE S2E0292E0n/e 52 E S2S1dVdS E0 q4rrdE02qdE e4 r2r0rE r dE0r R0 r RE r RdE0解1 1 EdVdS )01krE r 4r kr4r dr r224000kr2E r e4r0r )1kRE r 4r kr4r dr r224000kR2E r e4r0解2 q V kr4rr2)1 kr4rdrkr22E r re re4r42r000r )1 kr4rdrkR2R2E r ee4r4 r2r2r000 )解 E e21n0en 1xeE22x rn220 xE E E e212x rn220 x E x 1E e

33、21r /xn220e2n0 x r E E P E E .1212證 E r0Er ，E r21200 E E E r r12120r ar12E a0 r q EEdS q/dS Er2 0解 r E4r q/20q0E 0r R 1 1Q r R33 q R r R 11R R123321 Q r R33 E 故114 R R r2323201R r R Q 231QE 14 r320r R Q Q 312Q QE 124 r420r R 3QE E E 24 R432030E EdSE2rLq解 E2 q/0q 0r R ，1E 01 q LR r R ，12E 2 r20 q 0r

34、RE 03LE 2 r 2 rL 000 Q Q Q 132 W Q Edl20E Q Q 13 WQ V V QV202 0V Q Q O013解1 Q 1QQQQ0242d34 d12120011Q Q Q4423Q Q y 13QE E E y1y3y2 d y 3/2220Q Oy 2Q1Q2WQ E dl Q dy42 2y8 d2 2 d y 3/200001Q Q解2 Q Q O4213QQQV 134 d 4 d 2 d0000Q O2Q2WQV 2 08 d0 解 rU 12rE dr ln222r1r10E er 2 r0 解 qqpcosV V V 4 r 4 r 4 r

35、P20 0 0p0oV 2.2310 V3 若 若 若4 rP20pcos45oo90oV 1.5810 V34 rP20pcos90oV 04 rP20 R 1 qV 4 R02R3解 R q 111 qV 36V14 R101R 2Rq q 32121V 57V142301i0 02000V E dl xx00 xax0a0 x0V E dl ppQV 4 r0QV 4 R0R 解1 E 0r R11QE eR r R14 r22r120Q Q E e2r R14 r32r20由電勢V ElrR 1r1V 1 E dl R2 E dl E dl1123R2Q 1 1 Q Q01124 R

36、R4 R01202QQ124 R 4 R0102R R 12rV R2 E dl E l223R2Q 1 1 Q Q1124 r R4 R0202QQ124 r 4 R002R 2Q QV E dl 124 r33r0 R21210121QQV 1120102R R 12QQ122002R 2Q QV 1230 QQ 11121201021E dS V0babaV a bR 時(shí)20 rE r 得20R 時(shí)E2rl R l/20 R2E r 得2 r0R 時(shí) rV r Rdr R r2224r00R 時(shí)R2dr R2Rr V r Rln2 r2r00Vr 解 1 2rdrdV 4r x220P

37、rdrRV R x x2222r x02200 dVdx 1xE i i2R x220 x x VVE 5607Vm-1qR CR 28qV E 1695V4 x0q5649 Vm4 x20R 解 E 2 r0RU 12RE dl ln222R1210R2U /ln2.110 Cm812R1012 r E 7475Vm12 r0 eV 4 r0Ek1 e2E eV 4 r 2RK02R03E kTK 2解 1 e2E eV 7.210 eV54 r 2RK02R01E v2E /m 1.210 m sv 0由2 0712K0k0 3E E kT2K0K2ET K0 5.610 K得9k 解 E

38、 qUm8.9810 Kg4LL E 3000kWh 1.0810 J100E qUn2.8EE00 qV 4 R xl/2220 x 解 x qqVV 24 R xl/24 R xl/222200 當(dāng)xlxRqqV V 4 R x lxl /44 R x lxl /422222200q 11 4 x1l/x1l/xl0q l . 11. 4 x2x2x0ql4 x20在xR Oxy 電勢差U U ABAB11 U U V V22ABABRBR 解 QRV V4 R AR00QR 2RV 4 r r B000 1R U V V 26ABRB0 解 1 e2E 4 rp0e2W E 27.2eV

39、4 rp0 e / 4 r v /r222012e2E v 28 rk0e2E E E 8 rkp0E E 1360 q qqOq S 有1q S1 E dS E dS 0iS0iE E q d q q qF dbcd4 r20q dq qq qbcbc R F 解 E 4.8 J17epE E E 4.8 J17ekekep2Ev1.03 ms7ekmm E e2 rr0RV RE dr ln222 r2R1R100VE e reR2 RrRln201R11F E e N14r QV 若4 R002QV 若4 R002V E lppVR R02解 E20r R 1qE r R R 2220E

40、 r R 2220r R qQV E dl 1 E dl R2 E dl E dl 1123r0102R R 2qQV E dl R2 E dl E dl 223rR2002R 2qQV E dl 33r0r R ）qQ01022qQV 4 r 4 R2002R）2qQV 4 r30qQV V 4 R 4 R100201得qQV V 4 R 4 R100201r R E 0 V V；110R R 2RVRQRV (r RQE V 1 0；1 011r4 R rr4 Rr22220202R 2RV (R RQE 1 021r4 R r32202 QA Q Q Q ABA Q 接A q Aqq Q

41、 AAAqqq QV 0AAAA4 R 4 R4 RA010203V q 的AA解 qQQ QV 5.610 VA3AAA4 R 4 R4 RA010203Q QV 4.510 VB3A4 RB03 q AqqQ qV 0AAAA4 R 4 R4 RA010203Q qV AA4 RB03RRQq A2.1210 C12A8RR R R RR122313 V 0AV 7.2910 V2B d E Sd q/ 0E S0得 234S SS23E S0q S S 0得2323123140000 可解 S Q12 SQ34 014 023Q 2S1234QE 2 S0QdU 2 SAB0 014QS

42、 23QE S0QdU S0AB OqV 4 R0R qqV 4 R 4 Rs00解 為qqqQV 4 r 4 a 4 b000 Vq qqOdqsV4 R0OqV V4 r00V q0解 qqq1sq0sV 4 r4 R 4 r 4 R0000Rq dq qr 解 R R mR 65121 6RRC 44.58102F12R R021 解 d RU lnR0d RdC /ln /lnURR00C1012F 解 1 1 C S d d00Cd2d d d 0.152 mm0 mind C Smin000 U RE l211E2rL L解 0E 2 r0r R 1E 12 R01R11U R2

43、E r R2dr RE ln22 rR1110E 2.010 m 當(dāng)61 R R 112U V3 解 r SC r 0 1.5310 F9d U VQ C8Q 1.8410 Cm8-2S0 1 1 1.8310 Cm4-200r UE 1.210 Vm5d DdS q E D/ 0 r0V Edlrp解 r aD 4r 0r R21D 0 E 0；11D 4r QR r R d22QQD E ；4r4 r22220 rD 4r Qr R d23QD QE ；4r4 r32320 rr r r 1D 0 E 0；rr11r 2QQD 3.510 CmE 8.010 Vm82；214r4 rr2r

44、22220 r 2r 30QQD 1.310 CmE 1.410 Vm82；214r4 r0 2r2r2333 r 3QV E dr 360V4 r33r10r 2RdV E dr E dr223rQRdQ2Q 4 r 4 Rd 4 Rd0 r 2480V0 r0r 1RdV E dr E dr123RQRdQQ 4 R 4 Rd 4 Rd0 r540V0 r0 0 1Q P 1 E r 4 Rdnr0n2r 1Q P 1.610 Cm 82r4 Rdn2r 1Q P 1 E r4 Rnr0n2r 1QP 6.410 Cm82r4 Rn2rE 9.810 V/m 解 6 0 rU 5.1 V

45、2 DE ， F DE 0 r0 解 QD 4.510 Cm52S0DE 2.510 Vm61 0 rP D E 2.310 Cm510 D0 rDE ， P D E P 0解 S D2rL LD得D e2r rDE e 2 r0 r0 r 1 P D- E - e 2r0rr Q QV 12C C12Q Q Q120Q Q 12解 C 2 R10C 2 R20 rC1Q Q Q 0.510 C16C C 110012rCQ Q Q 1.510 C2r 1r6C C20012EE0rEE0r 解 C CC C C C23C C231123d 2d 2dC C C 間123231C C1 C C

46、1 解 C C C AC12C C C 8CD341111CCCCABACCD5C ABQ Q Q Q ACCDDBABCU U 4VABCACABACCU U 6VABCCDABCDCU U 2VABCDBABDB d QQ d U S0 S0 rE Q d U S0解 SC 0d0 SQ 0Ud0E U/d0 C 為1 QQ S d C Q/ 0r S S d 100 rr SUC CU 0 r d 11rQUE1 1 S d 0 rrQUE 1 1r S d 0r SC 0d 2 SQ U0d 2E 02UE d 2E 0 解 SC 0 rd d r d C S Sd d 0 r00 r

47、r0 r 1Cr 1r 1Cr SC 0d d0 Sd d 0C0 D XC L XC 而12C C C QCUQ僅12C Q 證 2 XC 0 rD1lnd 0 rDlnd2 L XC 1 2 X 2 L XC C C 0 r0 r XDDd12lnlnd 2 L2 L ； 00rDDlnlnddQ QXQ E 7bE b解 U E d Vmaxb 1W CU 9.0310 J2max32e E 2 R 00E E6bE b解 E 2.6610 Cm52max0b 2 RE max01bR1E R r R2 r rm210E 011 R E22w E 21b22r2m0m01RW w2rdr

48、 R Edr222rm01bR1R2RW R E ln 5.7610 Jm2241m01b1 解 E U/d Vm61E 3.010 VmE E61 ，bb VE 3.210 Vm 61r d rEE bE V/ Vm71bE 10Mm1，EEb 解 E E 1810 Vm61bU md U /E 2.2210 m4mb CdS 0.42m2 0 1 解 1Q2w E 222 Se020d Q d2W wV 2 See0 F F eQ d2A F r QEd 2 Se0 nr 1 RnR 2r r B Sm He II/cNecI l解 IlN 410ec mM/NN AAn /mj vmdv

49、d8kTv mek m e解 nN /MAj mv j /nej M/N e4.4610 ms41dmmA T v1 8kTv m2.42108vdde r j I/2解 r j I/2 m2 解 度 I2B 3/2004 2R2 RR24 2I A90 Obef0Idlr0B B 0I IOfa12 I l I lB 0 1 1 B 0 2 2,4r4r1222I I R l 12I l I l1 12 2 O解 O I l I lBB B 00 1 120 2 224r4r12 BOB0ilr 0解 OOO I08RB 0B 0 I IB 002R 2R0B 0 I I I I IB 00

50、0004R 4R 4R 2R 4R0B 0 B04R IB0O4ROOO解 I I I I IB 0i 0k i 0k k 0i 0k4R4R4R4R2R0 I I I I 1 1i IB i k000004R4R4R4R 4R03 I I IB 0i0j0k8R4R4R0 可將半圓柱面分割成寬度dIRdOO解 Il/RdBdI0ROxyl O B dBsin 0y I0 IB dsin dsin 02R R0 Rx20 IBB 0 Rx2B Ox OxB0dxd Bd Bd B222000) x2x2x2證 IR202R d/2x 3/2IR20R d/2x 3/2 B2222dB x IR

51、3 d/2x3 d/2x2 00dx2R d/2x 5/2R d/2x 5/222222dB x 3 IR4 d/2x4 d/2x R222 200dx22R d/2x 7/2R d/2x 7/22222B0由由，x d Rdxd B20dx2x0 B dB 01 d B 02 B x B 0 xx .2dx2 dx2dB 0dxd B 02x 0；0Odx2d R x BSS l BS 0 ldx2x 解 l Il d dldxln200212x2 dd1 解 I0I22R202R2r ， II I0R s/ 1.7810 m 3 I5.610 T03R r I B l011r2R1021

52、Ir021R r R1220 IB 0r2R r R23 r R 22 B 2r I I R R302232 I R r22B 032r R R32232r R3 B r I I 040B 04 r d B2B lI0dl B解 B2r I0r R1B r 01B 01R r R21B r 20 NIB 02r2r R2B r 03B 03R R R R 21121 R R R221 NIB02R r IrB r 02R2 B r dSS B rS解 Ir I Rdr 002R420 OOOOdOO d 解 d Bl Bl Bl Bl BllabcdbcdaB B B B 1234Bl 2ld

53、 I lBl 2 01B j201j 解 ，2 0 x j jB i i ji00220 j jB 0i -0i 022vBFv B 解 FLL 因vBF B1016NLG m g 1.64 N16pF /G10L 解 F vBq v2RmkyxR R y 2.98 m 223 E jC E vBH j v證 E j CE vBHE vBHE /E v HC qvBqEH解 EUv H0.63m/sHBdB 解 F IBlp JBlSSpJ 63.3810 A/m2Bl 解 pv kg21p2E 2.35keV2mk 解 v3eB1v23meB22 F F 12F F 3434 I I l0

54、1 24 I I l I I l30 1 20 1 22d2 d b34 IB,02dF BIBC UCUE F 2 dE0F F 0BE解 I2F BI B02dC U22F E2 d0 f 0由 fBE IC U22202d 2 d0CUI 4.510 A3 00 N W9 12 j0B01B B j21001B B j2200B0SF lB0解 1B B j20101B B j20201 B B B20121 j B B021B 0 dF jddyB1 jB B B022dSddy20210 解 m N 1.5610 Am72e I m/S A3 La h/20eei T 2a /v0

55、iB02a0解 hv2ma0ehei 2ma4 ma2200 i heB12.5T002a 8 ma2200 2rdrdI rdrTP r I2B 02r x 3/222B dBx P mr I2x m rdI解 x r dr R 2x322BR0 02x22r x 3/2x R20222rdr2R 2x22xBR002 (r x )222 3/2 x R022m1mRr dr R3440B mOx dl H2r HH dl If BH M 1 H，r解 H2r Ifr R1I I r2Rf21得IrH 2R121 r IrM 0 B ，02R1121R r R21I If得IH 2r2 rI

56、 I M 1B ，0rr2r2r2R r R32 I I I r R22 R Rf23223得 I R r22 H 32r R R32232 I R r22 M 0 B ，03r R R332232r R3 I I I 0f得H 0 M 0 B 0，444I Mrs 由 I M R R 1Isi211r I M R R 1Ise222r1 r R r1r R 2 VN NMA0N A矩m0 M mB0解 SLN NMA0SLmNm N m 7.85Am2M0A00 M mB Nm0 I NIH LB/SB H0r B/ Hr0NI/L /S和 解 L r4.78103 NIS0 di1dt M

57、21 ；21i2dt M1212 ddtddtN 解 d 2.51 cos100t Ndtt s V當(dāng)2ddt BSB SB B 12B B(x)B x d SxSydlE MdtM I Id BdS B dS B dS ddxddx 002 xd2x12 Id Id Id 3ln 2d d 2ddxdx 0002 xd2x24ddddt d 3 dIE 0ln2 4 dt dI 3 ln024 d 3M 0lnI2 4dl dtdIdt d 3 dIE Mln02 4 dt BS f SdE vdt dtd 解 NBS0 NBS21NBSqR R R Rii q RRBi 0.050T則NS 解 N ISN IS NBS NBS ,002r2r112212 N IS1 1 E 0 1.1110 V8t2t r r12 dE dtddtll E vB ll v OPR O或P 距 1 2Rx R2 B2即ddt E RvBOPP o/2v P OPddtBSE E 即E E EOPPOE E RvBOPPO A B OA OB O 1 E ABvB dl L-rlBdl lB L2r2AB-r1 U E lB L2r2ABABL r A B 中11 E Br E B Lr2,222OA