411_exam.final.sample.soln

411_exam.final.sample.soln - Physics 411 Final May 2001...

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Unformatted text preview: Physics 411 Final May 2001 SAMPLE Name__________________ 1.________(80 points) 2.________(35 points) 3.________(35 points) total .________(/150 points) __________________________________________________________________________________ DÝr Þ= PEÝr Þ 4 6 DÝr, t Þ = 4^k 1 O o _ 0 Ýr, t Þ ßD 1 Ýr, t Þ ? D 2 Ýr, t Þà 6 n = aÝr, t Þ 4 6 BÝr, t Þ = 0 /BÝr, t Þ /t k 2 J /DÝr, t Þ + 4^k J J r, t 4 × H Ýr , t Þ = Þ 2W 0Ý o k1 WoOo /t 4 × EÝr, t Þ = ?k 3 ßB 1 Ýr, t Þ ? B 2 Ýr, t Þà 6 n = 0 n ×ÝE 1 Ýr, t Þ ? E 2 Ýr, t ÞÞ = 0 n × ÝH 1 Ýr, t Þ ? H 2 Ýr, t ÞÞ = FÝr, t Þ ®Ý r Þ = 1 4^O o Ýv XXX |_?rrÞ| d 3 x + ® o v r V k?k v ?Ýg?g v Þtà 4 6 JÝr, t Þ+ /_Ýr, t Þ = 0; /t x NÝgÝxÞÞ= > i NÝdg ? x i Þ i F = qßE +k 3 v × B à; NÝ k ? k v Þ NÝ g ? g v Þ = 1 | dx | x=x i Ý 2^ Þ 4 XXXX all space, time e ?ißr6 d 3 rdt df = dr 6 4fÝr Þ system esu (Electrostatic) emu (Electromag.) Gaussian Heaviside-Lorentz SI (MKSA) SI (MKSA) quantity Ampere Oo = Wo = Wo Oo dq Coul.s ?1 dt 10 7 4^c 2 ?7 DÝr Þ= PEÝr Þ. J k3 1 1 c c 1 1 1 1 c 1 c XXX contains r fÝr v ÞN Ý3Þ Ýr ?r v Þdx v dy v dz v = fÝr Þ B k1 1 c2 1 1 4^ 1 4^O o k2 1 c2 q statcoul. statcoul. statcoul. statcoul 1 1 c2 1 4^c 2 Wo 4^ gauss (G) 1 Coulomb (C) 1 ” V/ohm = V/I tesla ” 10 ?7 c 2 10 ?7 1 8.854 187 10 ?12 Farad/m 1.256 637 10 ?6 Henry/m 376.730 610 5 /c 2 6.582 10 ?16 eVs e2 ci (SI) 4^10 376.730 ohms (IÞ 1.0546 10 ?34 Js ÝSIÞ = 1 137.04 i e e2 4^O o ci 1.602 10 ?19 Coulomb 4.80 10 16 statcoul Ýesu, GaussianÞ Ýesu, GaussianÞ Ýesu, GaussianÞ 0.529177 Å 0.529177 10 ?10 m e2 2r H Bohr radius, r H e2 4^O o Ý2r H Þ ÝSIÞ = 13.605 eV page2 DÝr Þ = 1 4^ dÝ r Þ = 1 4^ P XXX _ 0 Ýr v Þ Ý r ? r v Þ 3 v d r + D 0 Ýr Þ, SI units |r ? r v | 3 4 6 D 0 Ýr Þ = 0 XX surface aÝr v Þ dS v + d 0 . SI units |r ? r v | _Ýr Þ= aÝr ÞNÝFÝr ÞÞ|4FÝr Þ | ! EÝzÞ = n a 0 2P d + Ýr Þ ? d ? Ýr Þ = ? 1 ^ d Ýr Þ P å nÝr Þ6ÝP 1 E 1 Ýr Þ ? P 2 E 2 Ýr ÞÞ= aÝr Þ; å nÝr Þ6ÝP 1 E 1 Ýr Þ ? P 2 E 2 Ýr ÞÞ= 0; dÝ r Þ = ? 1 P N¸0 ?a 0 z = d x , y , z || Ý Þ 2P lim aÝr, N ÞN = ^ d Ýr Þ; XX ^ d Ýr Þ Ýr ? rÞ 6 dS v = ? 1 P |r ? r v | 3 dÝ r Þ = 1 P v XX ^ d Ýr Þ dI 4 2 dÝ r Þ = ? 1 _Ý r Þ P 4 2 gÝr, r v Þ = N Ý3 Þ Ýr ? r v Þ; dÝ r o Þ = XXX _Ýr v Þ gÝr, r v Þ d 3 r v ?1 + F Ýr , r 0 Þ 4^|r ? r 0 | gÝr, r 0 Þ = XXX gÝr, r o Þ_Ýr Þ/Pd 3 r+ XX volume bounding surface ! ßdÝr Þ4gÝr, r o Þ ? gÝr, r o Þ4dÝr Þà6nÝr ÞdS d D Ýr 0 Þ = 1 P XXX G D Ýr, r 0 Þ_Ýr Þd 3 r volume + XX bounding surface dÝr s Þ4G D Ýr s , r 0 Þ 6 dS 4 2 G N Ýr, r 0 Þ = N Ý3 Þ Ýr ? r 0 Þ; XX bounding surface ! 4G N Ýr, r 0 Þ 6 nÝr ÞdS=1 dÝ r 0 Þ = 1 P XXX G N Ýr, r 0 Þ_Ýr Þd 3 r volume + d avg + XX bounding surface ! G N Ýr, r 0 Þ4dÝr Þ 6 nÝr ÞdS ! ß4G N Ýr, r 0 Þà 6 nÝr Þ = d L?D Ýr o Þ = 4F N Ýr, r 0 Þ + 4 ?1 4^|r ? r 0 | ! 6 nÝr Þ = 1 S tot XX bounding surface dÝr s Þ L?D 4G D Ýr s , r 0 Þ 6 dS; d L?N Ýr Þ = ? XX bounding surface G N Ýr s , r 0 Þ4d L?N Ýr s Þ 6 dS SÝr, t Þ = EÝr, t Þ × HÝr, t Þ; W ij = P W =1 2 W electron = e 2 /2a Q i = > C ij V j j=1 N Wi = P 2 C ii = XXX |4d i Ýr Þ |2 d 3 r volume XXX 4d i Ýr Þ 6 4d j Ýr Þd 3 r, volume Qi Vi , Vj= 0 > QiVi= 1 2 i=1 3 N > C ij V i V j i,j=1 N Ißf, 4f à = XXX FÝr, fÝr Þ, 4fÝr ÞÞd 3 r / NF NF ? ? V u Ýr Þ = 0 Nf > /x i NÝ/f//x i Þ > J J i=1 J page 3 4 2 ®Ý_, j, z Þ = 1/ r 2 /r _ //_ ®Ý_, j, z Þ + _12 //j 2 ®Ý_, j, z Þ + / r 2 /r ®Ýr, S, d Þ ? 1 L 6 L®Ýr, S, d Þ = 0 r2 1/ _ /_ 2 /2 /z 2 ®Ý_, j, z Þ = 0 W m (d)=A m e imd +B m e ?imd ; m/2 P m Ýx Þ = Ý1 ? x 2 Þ § P 0 Ý1 Þ = 1 : § L 6 L Y(S,d)=§(§+1)Y(S,d) /2 WÝdÞ = ?m 2 WÝdÞ /d 2 d dx Ý1 ? x 2 Þ F §,|m| Ýx Þ=A §,|m| P |§m| (x)+B §,|m| Q |§m| (x) d dx F |lm| Ýx Þ + ߧݧ + 1 Þ ? m2 1?x 2 àF |lm| ÝxÞ = 0.; x = cos S d m P § Ýx Þ dx m ,m³0 , P 0 Ý ?1 Þ = Ý ?1 Þ § , § P m Ýx Þ = Ý?1 Þ m P |§m | Ýx Þ, m < 0 § if m ® 0, P m ݱ1 Þ = 0. P m Ý?x Þ = Ý?1 Þ l+m P m Ýx Þ § § § P 0 Ýx Þ = 1 P 1 Ýx Þ = x P 2 Ýx Þ = P 3 Ýx Þ = P 4 Ýx Þ = 1 2 x 2 1 8 Ý3x 2 ? 1 Þ Ý5x 2 ? 3 Þ Ý35x 4 ? 30x 2 + 3 Þ 2l+1 Ýl?m Þ! 4^ Ýl+m Þ! x + 1 m/2 FÝ?§, § + 1, 1 ? m; 1 ? x Þ x ? 1 m/2 2 1 ? x m FÝm ? §, § + m + 1, m + 1; 1 ? x Þ P m Ýx Þ =constant x + 1 § x?1 2 2 @Ýa + nÞ@Ýb + nÞ n @Ý 1 ? c Þ FÝa, b, c; zÞ = Fn z n!@Ýc + nÞ @ÝaÞ@ÝbÞ @ÝaÞ = Ýa ? 1Þ! Q m Ýx Þ =constant § Y m Ý S , d Þ = Ý ?1 Þ m l P m ÝcosÝS ÞÞe imd l 2^ ^ Y ?m ÝS, d Þ = Ý?1 Þ m Y m ÝS, d Þ D l l Y m Ý^ ? S, 2^ ? d Þ = Ý?1 Þ l Y ?m ÝS, d Þ l l Y 0 Ý S, d Þ = 0 1 4^ 3 Y 0 Ý S, d Þ = 1 4^ Y 1 Ý S, d Þ = ? 3 1 4^ 1/2 1/2 X 0 X 0 sin S Y m v ÝS, d Þ D Y m ÝS, d ÞdSdd=N ll v N mm v l lv 5 4^ Y 1 ÝS, d Þ = ? 15 2 8^ Y 2 ÝS, d Þ = ? 15 2 4^ Y 0 Ý S, d Þ = 2 1/2 1/2 1/2 l > lK 0 > m=?l Y m ÝS v , d v Þ D Y m ÝS, d Þ = NÝd ? d v ÞNÝcos S ? cos S v Þ l l = 3 cos 2 S ? 1 2 2 sin S cos Se id cos S sin Se id 1/2 sin 2 Se i2d 7 1/2 5 cos 3 S ? 3 cos S 2 4^ 2 1/2 Y 1 ÝS, d Þ = ? 1 21 sin Sß5 cos 2 S ? 1 àe id 3 4 4^ 1/2 Y 2 ÝS, d Þ = 1 105 sin 2 S cos Se i2d 3 4 2^ 1/2 Y 3 ÝS, d Þ = ? 1 35 sin 3 Se i3d 3 4 4^ 4 2 ®Ýr, S, d Þ = ?4^_Ýr, S, d Þ; 4 2 ®Ýr, S, d Þ = ?_Ýr, S, d Þ/O units K l rl 1 < m ®Ýr, S, d Þ = 4^q > l=0 > m=?l 2l+1 r l+1 Y l ÝS, d ÞY m ÝS v , d v Þ D l > ————————————————————————————————————————————– Y 0 Ý S, d Þ = 3 page 4 K l GÝr, r v Þ = ?1 > l=0 > m=?l 4^ l 4^ r < 2l+1 r l+1 > Y m Ý S, d Þ Y m Ý S v , d v Þ D l l G D Ýr , r v Þ = ?1 4^ 1 |r?r v | + ?R/ r v r?ÝR 2 /r v2 Þr v , K 1 = > l=0 v! |r ?r z| 1 => K > l l=0 m=?l |r ? r v | dÝ r Þ = 1 P l 4^ r < l+ 2l+1 r > 1 Y m Ý S, d Þ Y m Ý S v , d v Þ D l l rl < l+ r> 1 P 0 Ýcos S Þ l XXX V _Ýr v ÞGÝr, r v Þd 3 r v + XX S ßdÝr vs Þ4 v GÝr, r vs Þ ? GÝr, r vs Þ4 v dÝr vs Þà 6 dS v l 4^ 2l+1 P l Ýcos L Þ = > m=?l Y m Ý S, d Þ Y m Ý S v , d v Þ D l l ——————————————————————————————————————————/ /2 1/ 1 /2 _ /_ _ /_ ®Ý_, j, z Þ + _ 2 /j 2 ®Ý_, j, z Þ + /z 2 ®Ý_, j, z Þ = 0 d 2 W k Ýz Þ = k 2 W k Ýz Þ; W k Ýz Þ = A k e kz + B k e ?kz Note that k can dz 2 2 d V m Ýj Þ = ?m 2 V m Ýj Þ;V m Ýj Þ = C m e imj + D m e ?imj dj 2 2 _ 2 dd_ 2 F m Ýk_ Þ + _ dd_ F m Ýk_ Þ + Ýßk_à 2 + m 2 ÞF m Ýk_ Þ = 0 be complex, pure imaginary or real. F m Ýk _ Þ = a m J m Ýk _ Þ + b m N m Ýk _ Þ H ± Ýk_Þ ¯ J m Ýk_Þ ± iN m Ýk_Þ m Ýk_/2 Þ m J m k_ ¸0 i @Ý m + 1 Þ N m k_ ¸0 i @Ýp ÞÝk_/2 Þ ?m /^. J m Ýk_ ¸K Þ i N m Ýk_ ¸K Þ i 1 2.405 3.832 7.016 2 ^k_ 2 ^k_ cos k_ ? sin k_ ? m^ 2 m^ 2 ? ? ^ 4 ^ 4 n m=0 m=1 m=2 m=3 m=4 m=5 5.135 8.417 6.379 9.76 7.586 8.780 15.700 Table of x mn where J m Ýx mn Þ = 0 : 2 5.520 3 8.654 11.064 12.339 10.173 11.620 13.017 14373 4 11.792 13.323 14.796 16.224 17.616 18.982 5 14.931 16.470 17.960 19.410 20.827 22.220 J m Ýx ÞN m Ýax Þ ? J m Ýax ÞN m Ýx Þ = 0 a Ýa > 1 Þ J vm Ýx ÞN vm Ýax Þ ? J vm Ýax ÞN vm Ýx Þ = 0 Ýa > 1 Þ X 0 _J m Ýk n v _ ÞJ m Ýk n _ Þd_ = 0.5 N n v n ßaJ m+1 Ýk n a Þà 2 where J m Ýka Þ = 0. X 0 _J m Ýk_ ÞJ m Ýk v _ Þd_ = k ?1 NÝk ? k v Þ I m Ý|k|_Þ = i ?m J m Ýi|k|_Þ I m Ý|k|z ¸0 Þ i 1 @Ým+1 Þ |k|z 2 ?1/2 m K and ; K m Ý|k|_Þ = ^ i m+1 H + Ýi|k|_Þ m 2 K m Ý|k|z ¸0 Þ i @Ým Þ 2 ^ 2|k|z 2 |k|z 1/2 m , m ® 0 ( i ? ln |k|z 2 , m = 0Þ I m Ýkz ¸K Þ i Ý2^|k|z Þ expÝ|k|z Þ K m Ýkz ¸K Þ i expÝ?|k|z Þ. ————————————————————————————————————————————– page 5 4 2 GÝr; r v Þ = NÝr ? r v Þ = K v 1 _ NÝ_ ? _ v ÞNÝz ? z v ÞNÝj ? j v Þ K GÝr; r v Þ = K ?1 4^|r?r v | 1 NÝ_ ? _ v Þ = _ v X kJ p Ýk_ ÞJ p Ýk_ v Þdk NÝz ? z v Þ = 21^ X e ikÝz?z Þ dk = ^ X cosßkÝz ? z v Þàdk, ?K 0 0 v K . NÝj ? j v Þ = 21^ > m=?K e imÝj?j Þ = 21^ á1 + 2 cosßmÝj ? j v Þàâ ———————————————————————————————————————————— K l GÝr, r v Þ = ?1 > l=0 > m=?l 4^ l 4^ r < 2l+1 r l+1 > Y m Ý S, d Þ Y m Ý S v , d v Þ D l l + > ml= ?l Y m ÝS, d ÞY m ÝS v , d v Þ D l l GÝr; r v Þ = 4^ > l=0 GÝr; r v Þ = GÝr; r v Þ = 1 2^ K b 2l+1 ?R 2l+1 > Ý2l+1 Þ R 2l+1 ?a 2l+1 < Ýrr v Þ l+1 b 2l+1 ?a 2l+1 K > m=?K e imÝj?j Þ X 0 k g m Ýz; z v ÞJ m Ýk_ ÞJ m Ýk_ v Þdk v K 1 2^ K K > m=?K e imÝj?j v Þ X 0 k ?1 expÝkz < Þ expÝ?kz > ÞJ m Ýk_ ÞJ m Ýk_ v Þdk 2k GÝr, r v Þ = > lm X k 2 g l Ýk Þj l Ýkr Þj l Ýkr v ÞY m ÝS, d ÞY m ÝS v , d v Þ D dk l l 0 ®Ýr Þ = 4^ > l=0 > m=?l K l 1 2l+1 K Y m ÝS, d Þ XXX _Ýr v Þ l r l< Y m ÝS v , d v Þ D r v2 sin S v dS v dj v l l+1 r> —————————————————————————————————————————— 4 2 fÝr Þ + k 2 fÝr Þ = 0 r2 K d2 dr 2 fÝr Þ = > l=0 > m=?l M l Ýr ÞY m ÝS, d Þ l M l Ýr Þ = a l j l ÝkrÞ + b l n l ÝkrÞ j l Ýkr Þ = ^ 2kr K l d M l Ýr Þ + 2r dr M l Ýr Þ ? lÝl + 1 ÞM l Ýr Þ + k 2 r 2 M l Ýr Þ = 0. ^ X 0 r 2 j l Ýkr Þj l v Ýk v r Þdr XX Y m ÝS, d ÞY m v ÝS, d Þ D dI = 2kk v NÝk ? k v ÞN ll v N mm v l lv K2 l > lK 0 X 0 k j l Ýkr Þj l Ýkr v Þdk > m=?l Y m ÝS, d Þ D Y m ÝS v , d v Þ = ^ NÝr ? r v Þ l l 2 = J l+ 1 Ýkr Þ 2 page 6 page 7 obtain the final answer. It is not necessary to show work, but some partial credit will be given for work shown on this page. Bold face denotes vectors, k is a constant vector. n is the surface normal Each part is 7 points, except as marked. _________________________________________________________________________________ Note that all these questions can be answered using the formulas on the cover sheets. __________________________________________________________________________________ a) Q Q |r1aŷ| dddz________ 2 ____________ 2 4a a Gr, r r d 3 r inside a grounded sphere of radius R, what are the boundary 1. Fill in the blanks with the best, simplest answer. Carry out all operations to all space b) If r 1 r R conditions satisfied by Gr, r ? ___GRr, r Gr, Rr 0___________ c) Which of Maxwell’s equations gives rise to the continuity of the normal component of Br, t at the interface between two materials? ______ Br, t 0_________________ d) A spherical conductor with radius R has a surface charge density Q . What is the discontinuity 4R 2 in the tangential component of the electric field across the spherical interface? _0 (tang E always cont.)______ e) Q total all space fx,y,z0 dS fx, y, z|f|d 3 x __________Q total ________ f) The charge density inside a closed surface is zero and the electrostatic potential on the surface is given by r s Q Q Q r P cos . Inside the surface r ___ 3 rP 1 cos 3 z________ 3s1 a a a Q g) What is the electric field inside the surface described in part f)? __E 3 z_______ a h) Write down the most general form for r in spherical coordinates if 2 r 0 l ___r, , l0 ml A lm r l B lm r l1 Y m , _ l i) For a system of three conductors with potentials V i the charge on the third conductor is given by Q 3 3V 1 5V 2 7V 3 coulombs.. What is the capacitance of the third conductor ? _____7Farad_____ j) The charge density for a point charge, Q, located at r 0 is __Qr r 0 ___(3 points) k)One can use the variational method to determine the capacitance of a coaxial cable. In this approach, chooses a general form for the electrostatic potential, r between the two concentric conductors and imposes the condition that 0 on the outer conductor and __1__ on the inner conductor. Then one minimizes 2 W _ 1 i,j1 C ij V i V j _______ 2 l) In part (k), what is the expression for the capacitance, C, of the system in terms of the symbols given in (k)? _______C 2W______________ one page 8 2 A surface charge with density, Qr a/2 r3 cos is surrounded by a grounded conductor a with radius a. (15 pts. a). Give an explicit expression for the Green’s function for this problem. Gr, r 1 4 1 rr a/ r ra 2 /r 2 r (20pts, 2b) Find the electrostatic potential inside the sphere. Show all work. . r 1 1 r’ a r Gr, r d 3 r S r s Gr, r s Gr, r s r s dS r’ a Qr a/2 r3 cos Gr, r r 2 sin dr d d 0 0 a a Expand cos in Y m , : l 1 1 Q a/2 cos 4 3 3 1 rr a/ r ra 2 /r 2 r sin d d 1/2 Let a 2 /r 2a cos Y 0 , 1 r 1 3 Q a/2 [Y 0 , 3 1 3 1/2 3 4 1 rr (see table and do by inspection) 2 r2ar a 4 41 d and where r a/2 a 2 /r 2 r 4r r 2ar Next, expand the 1 rr and a/ r ra 2 /r 2 r in terms of Y m s. : a/r 2 l 1 rr l l 0 ml 4 2l1 r a/2 l1 a/2 r l substitute in r r 1 a/2 l rl r a/2 2 2a l1 2a rY m , Y m , l l l 1 r 2 r2ar and Q1 8 1 4 l l 0 ml Y m , 241 43 1/2 Y 0 , Y m , d 1 l l l r a/2 l1 a/2 r l Carry out the integration over solid angle l r 1 Q 1 41 l0 ml Y m , 241 4 1/2 m0 l1 l l 3 8 r a/2 l1 a/2 r l a/2 l rl r a/2 2 2a l1 2a r l 1 r a/2 l rl r a/2 2 2a l1 2a r l 1 r page 9 For r a : 2 r Q Q r r r r Y 0 , 4 1/2 a/2 2 2 2a 2 cos a/2 2 2 2a 2 1 3 8 3 8 3 7Q r cos r r a 16 3 a 2 2 a ra For 2 Q a/2 r r cos r 2 2 2a 2 8 3 Q a cos 1 r 3 2 2a 3 8 3 r 2 r Q a cos 1 16 3 r 2 r3 a3 Note that r 0 when r a page 10 3. A grounded right circular cylinder of radius b has it axis coincident with the z axis and its ends at z 0 and L. (a) (15 points) Write an explicit expression for the Dirichlet Green’s function for this problem? The Gr, r is given by the first form in Jackson problem 3.23.(cover page 6): Gr, r 1 b m n1 x mn b e x mn J m1 x mn 2 sinh x mn L b im J m x mn b Jm sinh x mn z sinh b x mn L z b Note that k x mn where J m x mn 0. The n1 is over all the zeros of the J m . Gr, r 0 on all b the boundaries of the cylinder. (b)(20 points) Using the above Green’s function find the electrostatic potential energy inside the cylinder if a point charge, Q, is located at L z,. Show all work. 3 r 1 inside cylinder r Gr, r d 3 r cylinder surface r s Gr, r s Gr, r s r s dS The surface integrals are all zero, since the cylinder is grounded and Gr, r 0 on S. r 1 1 3 The integral is done immediately using the r L z 3 Q r L z Gr, r d 3 r inside cylinder r Gr, r d 3 r to give r Q G r, L z 3 r Q1 b m n1 x mn im J m Jm b e x mn J m1 x mn 2 sinh x mn L b x mn b page 11 sinh x mn z sinh b x mn L z b where 0; 0; and z L 3 r Q1 b m n1 im J m x L z e sinh x mn z sinh mn x mn L b b 2 x mn J m1 x mn sinh b x mn b J m 0 J m 0 m0 so the sum over m reduces to one term, m 0: xn J 0 b x L z Q r 1b n1 sinh x n z sinh n b b x n J 1 x 2 sinh x n L b where x n are zeros of J 0 x , that is J 0 x n 0 for z L : 3 xn J L 0 b Q1 x n z sinh x n L 3 r b n1 sinh b b x n J 1 x 2 sinh x n L b or z L : 3 r Q1 b n1 J 0 xn b x n J 1 x 2 sinh x n L b sinh x n L sinh b3 x n L z b ...
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