411-exam.final.sample

411-exam.final.sample - Physics 411 Final May 2002 SAMPLE

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 411 Final May 2002 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 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 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 ?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 > ————————————————————————————————————————————– ^ ^ X 0 sinÝmzÞ sinÝm v zÞdz = ^ N mm v X 0 cosÝmzÞ cosÝm v zÞdz = ^ N mm v 2 2 Y 0 Ý S, d Þ = 3 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 ! |r ?r v z| 1 => K > l l=0 m=?l |r ? r v | dÝ r Þ = 1 P l 4^ r < l+1 2l+1 r > Y m Ý S, d Þ Y m Ý S v , d v Þ D l l rl < r l+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 d2 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. ————————————————————————————————————————————– 4 2 GÝr; r v Þ = NÝr ? r v = 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 GÝr; r v Þ 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^ 1 2^ 1 2^ 1 2^ K > m=?K e imÝj?j Þ X 0 k ?1 expÝkz < Þ expÝ?kz > ÞJ m Ýk_ ÞJ m Ýk_ v Þdk 2k K K mn mn mn > m=?K e imÝj?j v Þ > n=1 xa C mn expÝ xa z < Þ expÝ? xa z > ÞJ m v K K x mn D sinhÝ x mn z Þ sinhÝ? x mn z ÞJ > m=?K e imÝj?j Þ > n=1 a mn a< a>m K K n^ n^ zÞ sinÝ n^ z v ÞÞI m n^ _ < imÝj?j v Þ > m=?K e > n L E mn sinÝ L L L v K x mn _ J x mn _ v m a x mn _ J a mn _ v x m a a K m m^ _ > L 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 r l>+1 —————————————————————————————————————————— 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 K l > lK 0 X 0 k 2 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 ®Ýr Þ = > k > m=?K C km expß±kzà > m=?K e imj ßC km J m Ýk_Þ + D km N m Ýk_Þà ®Ýr Þ = > k > m=?K C km expß±ikzà > m=?K e imj ßC km J m Ýik_Þ + D km N m Ýik_Þà K K K K 1. Fill in the blanks with the best, simplest answer. Carry out all operations to 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. __________________________________________________________________________________ Q 4 1 à_d_djdz=____________________ 4^a 2 |r?aŷ| b) If dÝr Þ = ?P1 XXX v GÝr, r v Þ_Ýr v Þd 3 r v inside a grounded sphere of radius R, what are the boundary conditions r ²R satisfied by GÝr, r v Þ? _____________________________________________________ c) Which of Maxwell’s equations gives rise to the continuity of the normal component of BÝr, t Þat the interface between two materials? _______________________ a) XXX all space 4 6ß d) A spherical conductor with radius R has a surface charge density a= e) XXX Q total all space Q . What is the discontinuity in the 4^R 2 tangential component of the electric field across the spherical interface? __________________ XX fÝx,y,zÞ=0 dS NÝfÝx, y, zÞÞ|4f|d 3 x = __________________________________ f) The charge density inside a closed surface is zero and the electrostatic potential on the surface is given by Q dÝr s Þ = 3 r s P 1 Ýcos S Þ. Inside the surface dÝr Þ =______________________________ a g) What is the electric field inside the surface described in part f)? ____________________________ h) Write down the most general form for dÝr Þ in spherical coordinates if 4 2 dÝr Þ = 0 ______________________________________________________ i) For a system of three conductors 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 ? _____________________ j) The charge density for a point charge, Q, located at r 0 is ______________________(3 points) k)One can use the variational method to determine the capacitance of a coaxial cable. In this approach, one chooses a general form for the electrostatic potential, ®ÝrÞ between the two concentric conductors and imposes condition that ® = 0 on the outer conductor and ® =______ on the inner conductor. Then one minimizes W = ________________________ l) In part (k), what is the expression for the capacitance, C, of the system in terms of the symbols given in (k)? __________________________ the 2 A surface charge density, aÝrÞ = QNÝr ? a/2Þ r3 cos S 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. (20pts, 2b) Find the electrostatic potential inside the sphere. Show all work. 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? (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 ...
View Full Document

Ask a homework question - tutors are online