Griffiths Problems Solutions

Welcome to the archival Web page for U.C. Berkeley's Physics 110A , Section 1, Fall 2001. Email to: (Prof.) Mark Strovink, strovink@lbl.gov . I have a research web page , a standardized U.C. Berkeley web page , and a statement of research interests . Many of the problem set solutions for this course were handwritten initially by Graduate Student Instructor Ed Burns . They were refined, typeset, and additional solutions were composed by this course's GSI Daniel Larson . Course documents : Most documents linked here are in PDF format and are intended to be displayed by Adobe Acrobat [Reader], version 4 or later (Acrobat will do a better job if you un check "Use Greek Text Below:" on File-Preferences-General). General Information including schedules and rooms. Course Outline . Problem Set 1 Solution Set 1 Problem Set 2 Solution Set 2 Problem Set 3 Solution Set 3 Problem Set 4 Solution Set 4 Problem Set 5 Solution Set 5 Problem Set 6 Solution Set 6 Problem Set 7 Solution Set 7 Problem Set 8 Solution Set 8 Problem Set 9 Solution Set 9 Problem Set 10 Solution Set 10 Problem Set 11 Solution Set 11 Midterm 1 Solution to Midterm 1 Midterm 2 Solution to Midterm 2 Final Exam Mark Strovink Professo r Particle Experiment Mark Strovink, Ph.D. 1970 (Princeton). Joined UC Berkeley faculty in 1973 (Professor since 1980). Elected Fellow of the American Physical Society; served as program advisor for Fermilab (chair), SLAC (chair), Brookhaven, and the U.S. Department of Energy; served as D-Zero Physics Coordinator (1997 & 1998). Research Interests I am intere sted in experiments using element ary particles t o test discrete symmetries, ab solute predictions and other f undamental tenets of the Standard Model. Completed examples include early measurement of the parameters describing ch arge parity ( CP ) nonconservat ion in K meson decay; est ablishment of upper limits on the quark charge radius an d early observati on of the ef fects of gluon radiation in deep inelastic mu on scattering; and estab lishment of strin gent limits on rig ht-handed char ged curr ents bo th in muon de cay a nd in pro ton- antiproton collisions, t he latter via the sear ch for production of right-handed W bosons in the D-Zero experiment at Fermilab. After the discovery in 1995 by CDF and D-Zero of the top quark, we measured its mass with a combin ed 3% error, y ielding (with other input s) a stringent te st of loop corrections to the S tandard Model and an early h int that the Higgs bos on is light. If a Higgs-like signal is seen, we will need to measur e the top quark mass more than an order of magnitu de better i n order to determin e wh ether that signal arises from the SM Higgs. Current Projects A continuing o bjective is to under stand better ho w to measure the top quark mass. Top quarks ar e produced most ly in pairs; each decays pr imarily to b + W . The b \u2019s appear as jets of hadrons. Each W decays to a pair of jets or to a lepton and neutrino. For top mass measurement the most importan t channels are those in which either one or both of the W \u2019s dec ay into an elec tron or muon . For the single-lepton final states, we developed in 1994-96 and applied in 1997 a new technique that suppr esses backgr ounds (most ly from sing le W produ ction) without bi asing the a pparent top mass s pectra. For the dilepton fin al states, where backgrounds and systematic errors a re lower but two final-state neutrinos are undetected rather than one, a likelihood vs . top mass must be calculated fo r each event. During 1993-96 we developed a new prescription for this calculation that averages over the (unmeasured) neutrino rapidities, and we used it in 1997 to measure the top m ass to ~7% accuracy in th is more s parsely populated channel. In both channels , further improve ments to meas urement techn ique as well as accumulation o f larger samples will b e necessary. While studying data from the 1992-1996 CDF and D-Zero samples that contain both an electron and a muon, we became awa re of three events that cannot easily be attributed ei ther t o top quark dec ay or to ba ckgrounds . Generally this is because the transv erse momenta o f the leptons (electro ns, muons, and neut rinos as inferred from transv erse momentum imbalance) are unexpectedly large. We anticipate confirmi ng data e.g. from the D-Zero run that began in 2001. Transverse mo mentum imbala nce is a broa d signatur e for new p hysics. F or example, in many sup ersymmetr ic models, R -parity conservation requires every superparticle to dec ay even tually to a li ghtest superparticl e that, like th e neutrino, can be observed only by measuring a transverse momentum imbalance. Reliable detection of this signature is one of the severest challenges fo r collider detectors. D- Zero\u2019s unif orm and highly segm ented uranium/liquid argon calorimeter yields the best performance achieved so far. Buildin g on that, we have developed a new approach to analysis of transverse momentum imbalance that, for a given efficiency, yields up to five times fewer false positives. Recently we have grappled with the long-standing problem of searching with statistical r igor for new phy sics in sa mple s that should be de scribable by Standard Mo del processes \u2013 when the signatur es for new physics are not strict ly predefined. We have identified plausible methods for performing this type of analysis, and have exercised them on D-Zero data, but the methods inv olve sacrifices in sen sitivity that we are still working to mitigate. Selected Publica tions S. Abachi et al . (D-Zero Collab oration ), \u201cSearc h for ri ght-ha nded W bosons and heavy W \u2032 in proton-antiproton collisio ns at \u221a s = 1.8 TeV,\u201d Phys. R ev. Lett . 76 , 3271 (1996). S. Abachi et al . (D-Zero Collaborati on), \u201cObs ervation of th e top qua rk,\u201d Phys. Rev. Lett . 74 , 2422 (1995). B. Abbott et al . (D-Zero Collaborati on), \u201cDirect measurement of the top quar k mass,\u201d Phys. Rev. Lett . 79 , 1197 (1997); Phys. Rev. D 58 , 052001 (1998). B. Abbott et al . (D-Zero Collaboration), \u201cMeasu rement of the top qu ark mass using dilepton events,\u201d Phys. Rev. Lett . 80 , 2063 (1998); Phys. Rev. D 60 , 052001 (1999). V.M. Abazov et al . (D-Zero Co llaboration), \u201cA quasi-model-indepe ndent search for new high p T physics at D-Zero,\u201d Phys. Rev. Lett . 86 , 3712 (2001); Phys. Rev. D 62 , 092004 (2000); Phys. Rev. D 64 , 012004 (2001). (510) 486-7087 strovink@lbl.gov University of California, Berkeley Physics 110A, Section 1, Fall 2001 (Strov ink) GENERAL INFO RMATIO N (25 Sep 01) Web site for this course: http://d0lbln.lbl.gov/110af01-web.htm . Instructors: Prof. Mark Strovink , 437 LeConte; (LBL) 486-7087; (hom e, before 10) 486-8079; (UC) 642- 9685. Email: strovink@lbl.gov . Web: http://d0lbln.lbl.gov/ . Office hours: M 3:15- 4:15, 5:30-6:30. Mr. Daniel Larson , 281 LeConte, (UC) 642- 5647. Em ail: dtlarson@socrates.berkeley.edu . O ffice hours (in 281 LeConte): Th 11:30-12:30, F 2- 3. Lectures: MWF 10:10-11:00 in 329 LeConte, and Tu 5:10- 6:30 in 329 LeConte. The Tu 5:10-6:30 slot will be used occasionally during the semester for the m idterm exams; for rev iews and special lectures; and for lectures that substitute for those which would norm ally be deliv ered later in the week . Lecture attendance is strongly encouraged, since the course content is not exactly the same as that of the text. Discussion Sections : W 4:10-5 in 409 Dav is, and Th 4:10-5 in 385 LeConte. Beg in in second week. Taught by Mr. Larson. You are especially encouraged to attend discussion section reg ularly. There you w ill learn techniques of problem solv ing, with particular application to the assig ned exercises. Texts : \u2022 Griffiths, Introduction to Electrodyamics (3 rd ed., Prentice- Hall, 1999, required). Get the fourth (or later) printing, which has few er typos. I feel that this text is well written and pedag ogically effective, thoug h its scope is modest and its problem s are som etimes not v ery phy sical. \u2022 If y ou are planning to attend physics g raduate school, it would be smart now to purchase Jack son, Classical Electrodynamics (3 rd ed., Wiley). Optionally , it can be useful in this course. Problem Sets : A required and m ost important part of the course. Elev en problem sets are assigned and g raded. Problem sets are due on Friday s at 5 PM, beginning in week 2. Exceptions : no problem set is due in the week preceding each m idterm exam ; the problem set that normally would be due on Friday of the week of the second midterm exam instead is due four days later, on Tuesday of Thank sgiving week (no other problem set is due on Thanksgiv ing week ). Deposit problem sets in the box labeled \u201c110A Section 1 (Strov ink)\u201d in the second floor breezeway between LeConte and Birg e Halls. You are encouraged to attem pt all of the problem s. Students who do not do so find it alm ost impossible to learn the m aterial and to succeed on the exam inations. Late papers will not be graded. Y our lowest problem set score w ill be dropped, in lieu of due date extensions for any reason. You are encourag ed to discuss problems with others in the course, but y ou must write up y our homew ork by yourself. (I t is straightforw ard to identify solutions that are written collectiv ely; our policy is to divide the score am ong the collectiv ists.) Exams : There will be two 80- m inute midterm examinations and one 3- hour final exam ination. Before confirming your enrollm ent in this class, please check that its final Exam Group 1 does not conflict with the Exam G roup for any other class in which y ou are enrolled. Please verify now that you w ill be available for the midterm exam inations on Tu 16 Oct (in 4 LeConte) and Tu 13 Nov (in 50 Birge), both at 5:10- 6:30 PM; and for the final examination on W 12 D ec, 8-11 A M. Except for unforeseeable emerg encies, it will not be possible for the midterm or final exam s to be rescheduled. Passing 110A requires passing the final exam. Grading: 25% problem sets, 35% m idterm s, 40% final exam. Departm ental regulations call for an A : B : C distribution in the ratio 2:3:2, with approximately 10- 15% D \u2019s or F \u2019s. However, the fraction of D \u2019s or F \u2019s depends on you; no m inimum num ber need be given. Physics 110A, Sec. 1 COURSE OUTLINE Fall 2001 (Strovink) Week Week Lecture Topic Problem Due 5 PM No. of... reference Set No. on... (Griffiths) 1 27-Aug 1.1.5, 1.3.2-1.3.6 Vector and tensor transformations, fundamental theorems 1.4-1.6 Curvilinear coordinates, Dirac delta function, theory of vector fields 2.1-2.2.3 Electrostatic fields, Gauss's law 2 3-Sep LABOR DAY 2.2.4-2.3 Electrostatic potential and boundary conditions 2.4, 2.5.1 Electrostatic work and energy, conductors 1 7-Sep 3 10-Sep 3.1.1-3.1.4 Laplace's and Poisson's equation, simple and relaxation solutions 3.1.5, 3.2.1-3.2.2 Uniqueness of solution, method of images 3.3.1 Separation of variables in Cartesian coordinates 2 14-Sep 4 17-Sep 3.4.2, 3.4.4 Ideal electric dipole and its field 4.1-4.2.1 Forces and torques on electric dipoles; polarization 4.3-4.4.1 Gauss's law in dielectrics, D , linear dielectrics 3 21-Sep 5 24-Sep 4.4.3-4.4.4 Energy in dielectrics, forces on dielectrics 5.1.1-5.1.2 Lorentz force law, particle trajectories in static fields 5.1.3 Current, forces on wires, current densities; charge conservation 4 28-Sep 6 1-Oct 5.2, 5.3.1-5.3.2 Biot-Savart law, divergence of B 5.3.2-5.3.4 Ampere's law and applications, static Maxwell equations 5.4.1-5.4.2 Vector potential, magnetostatic boundary conditions 5 5-Oct 7 8-Oct 5.4.3 Ideal magnetic dipole and its field 6.1.1-6.1.2, 6.1.4 Forces and torques on magnetic dipoles; magnetization 6.3, 6.4.1 Ampere's law in magnetic materials, H , linear magnetic media 8 15-Oct --- TBA (16-Oct) --- MIDTERM 1 (covers PS 1-5), in 4 LeConte 6.4.2 Ferromagnetism 6 19-Oct 9 22-Oct 7.1 Ohm's law, EMF 7.2.1-7.2.2 Faraday's law 7.2.3-7.2.4 Energy in magnetic fields, inductance 7 26-Oct 10 29-Oct 7.3.1-7.3.3 Maxwell's equations in free space 7.3.5-7.3.6 Maxwell's equations in matter, boundary conditions 10.1 Maxwell's equations for potentials; gauge transformations 8 2-Nov 11 5-Nov 8.1.1-8.1.2 Continuity equation, Poynting's theorem 9.1.1-9.1.2 Wave equation in one dimension, general solution, sinusoidal waves 9.2 EM waves in vacuum, energy and momentum 12 12-Nov --- VETERANS DAY (13-Nov) --- MIDTERM 2 (covers PS 1-8), in 50 Birge 9.3.1-9.3.2 EM waves in a linear insulator, reflection at normal incidence 14 19-Nov 11.1.1-11.1.2 EM fields of an oscillating electric dipole 11.1.1-11.1.2 Electric dipole radiation and power 9 20-Nov (22-Nov) --- THANKSGIVING 13 26-Nov 9.1.4 Polarization and angular momentum of EM waves; how to control 9.4.1-9.4.2 EM waves in a conductor, reflection at normal incidence 9.5.1, 9.5.3 EM waves in a coaxial cable 10 30-Nov 15 3-Dec --- Interference and coherence of >1 dipole radiator --- Radiation pattern from >1 dipole and connection to diffraction (Babinet) --- TBA 11 7-Dec 16 10-Dec --- (12-Dec) --- Final exams begin (12-Dec) 8-11 AM 110A FINAL EXAM (Group 1) (covers PS 1-12) 9/25/2001 Univ ersit y of California, Berk eley Ph ysics 110A F all 2001 Section 1 ( Str ovink ) Problem Set 1 1. Gri\ufb03ths 1.14 2. Gri\ufb03ths 1.16 3. Gri\ufb03ths 1.21 4. Gri\ufb03ths 1.33 5. Gri\ufb03ths 1.38 6. Gri\ufb03ths 1.46 7. Gri\ufb03ths 2.6 8. Gri\ufb03ths 2.16 Univ ersit y of California, Berkeley Ph ysics 110A F all 2001 Section 1 ( Str ovink ) Solution Set 1 1. Gri\ufb03ths 1.14 Under a rotation, the co ordinates y and z transform in to ¯ y = y cos \u03c6 + z sin \u03c6 and ¯ z = \u2212 y sin \u03c6 + z cos \u03c6 , so w e can inv ert these equations to \ufb01nd y =¯ y cos \u03c6 \u2212 ¯ z sin \u03c6 and z =¯ y sin \u03c6 +¯ z cos \u03c6 . Using the c hain rule: \u2202f \u2202 ¯ y = \u2202f \u2202y \u2202y \u2202 ¯ y + \u2202f \u2202z \u2202z \u2202 ¯ y = \u2202f \u2202y cos \u03c6 + \u2202f \u2202z sin \u03c6 \u2202f \u2202 ¯ z = \u2202f \u2202y \u2202y \u2202 ¯ z + \u2202f \u2202z \u2202z \u2202 ¯ z = \u2202f \u2202y ( \u2212 sin \u03c6 )+ \u2202f \u2202z cos \u03c6. Th us \ue001 ( \u2207 f ) y ( \u2207 f ) z \ue002 = \ue001 cos \u03c6 sin \u03c6 \u2212 sin \u03c6 cos \u03c6 \ue002 \ue003 ( \u2207 f ) y ( \u2207 f ) z \ue004 \u21d2 ( \u2207 f ) i = R ij ( \u2207 f ) j . So \u2207 f transforms lik e a v ector. 2. Gri\ufb03ths 1.16 The sk etc h of this vector \ufb01eld appears in Figure 1.44 of the text. (In fact, a whole discussion app ears in Section 1.5.1.) W e wan t to calculate the divergence of the v ector \ufb01eld v = \u02c6 r /r 2 =( x, y , z ) /r 3 . If y ou calculate the divergence in cartesian coordinates, the form ula for the div ergence is simple, \u2207 · v = ( \u2202v x /\u2202 x )+( \u2202v y /\u2202 y )+( \u2202v z /\u2202 z ), how ever, the deriv ativ es can get a little tiresome. Instead, I\u2019ll calculate the div ergence in spherical coordinates. A general v ector \ufb01eld in spherical co ordinates lo oks like v = v r \u02c6 r + v \u03b8 \u02c6 \u03b8 + v \u03c6 \u02c6 \u03c6 and the form ula for the div ergence is a little complicated (see the inside fron t cov er of Gri\ufb03ths): \u2207 · \ue00c v = 1 r 2 \u2202 \u2202r ( r 2 v r )+ 1 r sin \u03b8 \u2202 \u2202\u03b8 (sin \u03b8v \u03b8 )+ 1 r sin \u03b8 \u2202v \u03c6 \u2202\u03c6 Ho w ever, in our case, v \u03b8 = v \u03c6 \u2212 0, so the calculation isn\u2019t so bad. In fact, since v r =1 /r 2 , \u2202 ( r 2 v r ) /\u2202 r = \u2202 (1) /\u2202 r =0 . T h u s \u2207 · v = 0! This is surprising, b ecause the vector \ufb01eld certainly looks like it is div erging away f r o m t h e o r igin. The explanation is that the div ergence is zero everywhere exc ept at the origin; at the origin the ab o ve calculation fails because the vector \ufb01eld is unde\ufb01ned there. 3. Gri\ufb03ths 1.21 (a) ( A · \u2207 ) B =( A x \u2202 \u2202x + A y \u2202 \u2202y + A z \u2202 \u2202z ) B =( A x \u2202B x \u2202x + A y \u2202B x \u2202y + A z \u2202B x \u2202z ) \u02c6 x +( A x \u2202B y \u2202x + A y \u2202B y \u2202y + A z \u2202B y \u2202z ) \u02c6 y +( A x \u2202B z \u2202x + A y \u2202B z \u2202y + A z \u2202B z \u2202z ) \u02c6 z . This can b e expressed more succin tly using index notation and the summation conv ention: [( A · \u2207 ) B ] i = A j \u2202 j B i . (b) As in an earlier problem, the computation is a little easier in spherical co ordinates; how ev er, this time I\u2019ll use cartesian co ordinates for v ariety . F or the x comp onen t [( \u02c6 r · \u2207 ) \u02c6 r ] x = 1 \ue005 x 2 + y 2 + z 2 \ue003 x \u2202 \u2202x + y \u2202 \u2202y + z \u2202 \u2202z \ue004 x \ue005 x 2 + y 2 + z 2 = 1 r \ue006 x \ue007 1 r + x \ue003 \u2212 1 2 \ue004 1 r 3 2 x \ue008 + yx \ue007 \u2212 1 2 1 r 3 2 y \ue008 + zx \ue007 \u2212 1 2 1 r 3 2 z \ue008\ue009 = 1 r \ue00a x r \u2212 x r 3 \ue00b x 2 + y 2 + z 2 \ue00c \ue00d = 1 r \ue00e x r \u2212 x r \ue00f =0 . The other t w o comp onents are the same, just sw apping x for y or z . Th us all three comp onen ts v anish, so [( \u02c6 r · \u2207 ) \u02c6 r ]=0 . 1 (c) No w take v a = x 2 \u02c6 x +3 xz 2 \u02c6 y \u2212 2 xz \u02c6 z and v b = xy \u02c6 x +2 yz \u02c6 y +3 zx \u02c6 z . ( v a · \u2207 ) v b = \ue003 x 2 \u2202 \u2202x +3 xz 2 \u2202 \u2202y \u2212 2 xz \u2202 \u2202z \ue004 ( xy \u02c6 x +2 yz \u02c6 y +3 xz \u02c6 z = x 2 ( y \u02c6 x +0 \u02c6 y +3 z \u02c6 z )+3 xz 2 ( x \u02c6 x +2 z \u02c6 y +0 \u02c6 z ) \u2212 2 xz (0 \u02c6 x +2 y \u02c6 y +3 x \u02c6 z ) = x 2 ( y +3 z 2 ) \u02c6 x +2 xz (3 z 2 \u2212 2 y ) \u02c6 y \u2212 3 x 2 z \u02c6 z 4. Gri\ufb03ths 1.33 . W e w an t to test Stok es\u2019 Theorem for the function v =( xy ) \u02c6 x +( 2 yz ) \u02c6 y +( 3 zx ) \u02c6 z o v er the triangular region sho wn in the \ufb01gure. First let\u2019s calculate the in tegral of the curl o ver the triangle\u2019s area. \u2207 × v = \u02c6 x (0 \u2212 2 y )+ \u02c6 y (0 \u2212 3 z )+ \u02c6 z (0 \u2212 x )= \u2212 2 y \u02c6 x \u2212 3 z \u02c6 y \u2212 x \u02c6 z . Since the path around the outside is going coun terclo ckwise, the con ven tion is that the area element d a = dy dz \u02c6 x .T h u s e ( \u2207 × v ) · d a = \u2212 2 yd yd z . \ue010 ( \u2207 × v ) · d a = \ue010 2 0 \ue003 \ue010 2 \u2212 z 0 \u2212 2 yd y \ue004 dz = \ue010 2 0 \u2212 (2 \u2212 z ) 2 dz = 1 3 (2 \u2212 z ) 3 \ue011 \ue011 \ue011 \ue011 2 0 = \u2212 8 3 . No w w e calculate the path integral around the boundary of the triangle. v · d l =( xy ) dx +( 2 yz ) dy +( 3 zx ) dz . There are three segmen ts. T ak e the segment along the y -axis \ufb01rst. Here y go es from 0 to 2, but dx and dz are zero, and x = y =0 . T h u s v · d l = 0 so its in tegral is zero also. Let the second segment be the slanted side. Again, x = dx = 0, while z =2 \u2212 y and dz = \u2212 dy . In tra v eling up and left y g o e sf r o m2t o0 . S o \ue012 v · d l = \ue012 0 2 2 yz d y = \u2212 \ue012 2 0 2 y (2 \u2212 y ) dy = \u2212 (2 y 2 \u2212 2 3 y 3 ) \ue011 \ue011 2 0 = \u2212 8 3 . And the \ufb01nal segmen t is coming do wn the z -axis, where dx = dy =0 a n d x = y =0 s o v · d l = 0 and th us there is no contribution to the in tegral. So the con tributions from all three segments giv e \ue013 v · d l = \u2212 8 3 . Thus in t his case we ha ve demonstrated that Stok es\u2019 theorem holds, namely \ue012 S ( \u2207 × v ) · d a = \ue013 P v · d l . 5. Gri\ufb03ths 1.38 The divergence theorem tells us that the in tegral of the divergence o v er the v olume equals the in tegral of the vector \ufb01eld o ver the surface. So we need to compute t w o in tegrals in each case. (a) \u2207 · ( r 2 \u02c6 r )= 1 r 2 \u2202 \u2202r ( r 4 )=4 r , where I used equation (1.71). Now, \ue012 ( \u2207 · v 1 ) d\u03c4 = \ue012 (4 r ) r 2 dr sin \u03b8d \u03b8d \u03c6 = \ue00e \ue012 R 0 4 r 3 dr \ue00f \ue00b \ue012 d \u2126 \ue00c = r 4 \ue011 \ue011 R 0 (4 \u03c0 )=4 \u03c0R 4 . On the other hand, on the surface of the sphere, r = R ,a n ds o \ue012 v 1 · d a = \ue012 ( R 2 \u02c6 r ) · ( R 2 sin \u03b8d \u03b8 , d \u03c6 \u02c6 r )= R 4 \ue012 d \u2126=4 \u03c0R 4 . The tw o integrals agree. (b) \u2207 · (1 /r 2 ) \u02c6 r = 1 r 2 \u2202 \u2202r ( r 2 1 r 2 )=0 , s o \ue012 ( \u2207 · v 2 ) d\u03c4 = 0. On the other hand, on the surface of a sphere radius R , \ue012 v 2 · d a = \ue012 ( 1 R 2 \u02c6 r ) · ( R 2 d \u2126 \u02c6 r )= \ue012 d \u2126=4 \u03c0 . The tw o integrals don \u2019t agree! The reason is that \u2207 · v 2 =0 except at the origin, where it b ecomes in\ufb01nite. Thus our calculation of \ue012 ( \u2207 · v 2 ) d\u03c4 is incorrect. W e\u2019ll learn ho w to \ufb01x this using the Dirac delta-function. The correct answer is 4 \u03c0 , whic is what w e got the using the surface in tegral b ecause that metho d a v oids the problem at the origin. 6. Gri\ufb03ths 1.46 (a) Since the c harge is only at the sp eci\ufb01c p oint r \ue002 we will need to use a delta-function. \u03c1 ( r )= q\u03b4 3 ( r \u2212 r \ue002 ). The v olume integral is then \ue012 \u03c1 ( r ) d\u03c4 = q \ue012 \u03b4 3 ( r \u2212 r \ue002 ) d\u03c4 = q as it should b e. (b) The electric dip ole is just t wo di\ufb00eren t p oint c harges at di\ufb00erent places. \u03c1 ( r )= q\u03b4 3 ( r \u2212 a ) \u2212 q\u03b4 3 ( r ). (c) There is no c harge an ywhere except where r = R ,s o \u03c1 ( r )= A\u03b4 ( r \u2212 R ). W e need to integrate o v er all space to \ufb01nd A . Q = \ue012 \u03c1d \u03c4 = \ue012 A\u03b4 ( r \u2212 R )4 \u03c0r 2 dr = A 4 \u03c0R 2 .S o A = Q 4 \u03c0R 2 .T h u s \u03c1 ( r )= Q 4 \u03c0R 2 \u03b4 ( r \u2212 R ). 7. Gri\ufb03ths 2.6 The \ufb01rst thing to notice is that any horizon tal comp onents of the electric \ufb01eld will cancel because of the symmetry of the disk. So the resulting electric \ufb01eld will be in the \u02c6 z direction. Then w e just need to add up the con tributions of the z -comp onen ts due to ev ery p oin t on the disk. F or a generic point on the disk lo cated at a distance r from the center, the distance to the point P is \u221a r 2 + z 2 .T h u s t h e z -comp onen t of the E \ufb01eld at P due to that p oin t is 1 4 \u03c0\ue011 0 dq r 2 + z 2 cos \u03b8 where \u03b8 is the angle b et w een the line connecting P with the cen ter of the disk and the line connecting P with the generic element of c harge on the disk (see Fig. 1). So we 2 r \u03b8 z P Figure 1: Problem 7. see cos \u03b8 = z \u221a r 2 + z 2 . T o get the total E -\ufb01eld at P we need to add up these con tributions for every little elemen t of c harge on the disk. The amount of c harge in a small area element is dq = \u03c3 r dr d\u03c6 .T h u s E = \ue010 R 0 \ue010 2 \u03c0 0 1 4 \u03c0\ue01b 0 z\u03c3r d r d \u03c6 ( r 2 + z 2 ) 3 / 2 \u02c6 z = 2 \u03c0z\u03c3 4 \u03c0\ue01b 0 \ue010 R 0 rd r ( r 2 + z 2 ) 3 / 2 \u02c6 z = 2 \u03c0z\u03c3 4 \u03c0\ue01b 0 \ue014 \u2212 ( r 2 + z 2 ) \u2212 1 / 2 \ue015 R 0 \u02c6 z = z\u03c3 2 \ue01b 0 \ue007 1 z \u2212 1 \u221a R 2 + z 2 \ue008 \u02c6 z . Finally , it is alwa ys go o d to chec k the limiting cases of your results. F or R \u2192\u221e the second term v anishes and w e\u2019re left with \u03c3/ 2 \ue01b 0 whic h is the electic \ufb01eld due to an in\ufb01nite sheet of c harge. F or z \ue008 R at \ufb01rst glance the t w o terms in brack ets cancel, so we\u2019d get zero, whic h is true, but not so informative. T o get a more precise estimate of ho w the \ufb01eld falls o\ufb00 for large z we need to expand the square root. 1 \u221a R 2 + z 2 = \ue007 z 2 \ue003 1+ R 2 z 2 \ue004\ue008 \u2212 1 / 2 = 1 z \ue003 1 \u2212 R 2 2 z 2 + O \ue003 R 4 z 4 \ue004\ue004 Since z \ue008 R , we can ignore the R 4 /z 4 piece. Plugging the expansion back in to the formula for the electric \ufb01eld, w e \ufb01nd E \u2192 \u03c3\u03c0R 2 4 \u03c0\ue011 0 z 2 whic h is the \ufb01eld a distance z away f r o m a p o i nt ch a r g e with Q = \u03c3\u03c0R 2 . 8. Gri\ufb03ths 2.16 W e can use Gauss\u2019s la w with cylindrical surfaces to determine the electric \ufb01eld in eac h region. F ollowing the notation in Gri\ufb03ths, \u02c6 s is the radial unit v ector in cylindrical co ordinates. (i) F or s , w e imagine a small cylindrical surface of radius s and length \ue01f inside the inner cylinder. W e kno w b y symmetry that the resulting electric \ufb01eld m ust p oint radially out ward and is th us p erp endicular to the curv ed sides of our Gaussian cylinder. Thus \ue013 E · d a =2 \u03c0s \ue01fE ( s ). Gauss\u2019s law tells us this is equal to 1 \ue011 0 Q enc = 1 \ue011 0 \u03c0s 2 \ue01f\u03c1 . Equating these tw o expressions and solving for the electric \ufb01eld we \ufb01nd E ( s )= s\u03c1 2 \ue011 0 \u02c6 s . (ii) No w we imagine our gaussian surface to be b etw een the cylinders. The \ufb02ux of electric \ufb01eld lea ving the cylinder is the same as ab o v e, namely \ue013 E · d a =2 \u03c0s\ue01f E ( s ), while the other half of Gauss\u2019s law giv es 1 \ue011 0 Q enc = 1 \ue011 0 \u03c0a 2 \ue01f\u03c1 .T h u s f o r a , E ( s )= a 2 \u03c1 2 \ue011 0 s \u02c6 s . (iii) \ue013 E · d a =2 \u03c0s\ue01f E ( s )= 1 \ue011 0 Q enc = 0 because we are told that the whole wire is neutral, so the enclosed c harge on the central cylinder in cancelled b y the enclosed charge on the outer cylindrical shell. Thus for s>a , E =0 . The plot of the electric \ufb01eld as a function of s starts at zero, increases linearly until s = a , then it decreases along a segmen t of a hyperb ola un til s = b , and which point it abruptly drops to zero for all s>a . 3 Univ ersit y of California, Berk eley Ph ysics 110A F all 2001 Section 1 ( Str ovink ) Problem Set 2 1. Gri\ufb03ths 2.18 2. Gri\ufb03ths 2.20 3. Gri\ufb03ths 2.25 (c) only 4. Gri\ufb03ths 2.32 (a) and (b) only 5. Gri\ufb03ths 2.36 (a), (b), and (c) only 6. Gri\ufb03ths 2.39 7. Gri\ufb03ths 2.50 8. According to the Pro ca equations (a rel- ativistically in v arian t linear generalization of Maxw ell\u2019s equations accommo dating the p os- sibilit y of a \ufb01nite rest mass m 0 for the photon), Gauss\u2019s la w is mo di\ufb01ed to b ecome \u2207· E = \u03c1 \ue003 0 \u2212 \u03c6 ¯ \u03bb 2 , where \u03c6 is the electrostatic p oten tial and ¯ \u03bb \u2261 ¯ h m 0 c is the reduced (b y 2 \u03c0 ) Compton w a v elength of the photon. F ollo wing Williams, F aller, and Hill, Phys. R ev. L ett. 26 , 721 (1971), consider tw o concentric spherical p erfectly conducting shells of radii R 1 and R 2 , resp ectively , with R 2 >R 1 . Imagine that the inner sphere is isolated and that the outer shell is driv en b y an RF oscillator so that it has a potential (relativ e to \u221e ) \u03c6 2 ( t )= V 0 cos \u03c9t . In the mo di\ufb01ed form of Gauss\u2019s la w, make the follo wing approximation for the v alue of \u03c6 whic h app ears in the last term; this is a v alid ap- proac h b ecause the factor ¯ \u03bb \u2212 2 m ultiplying it is v ery small. The appro ximation is to set \u03c6 = \u03c6 2 ev erywhere within the outer sphere. Construct a Gaussian surface consisting of a third sphere at radius r , where R 1 2 . Consider the v olume in tegral of \u2207· E \u2212 \u03c1 \ue003 0 + \u03c6 ¯ \u03bb 2 within that surface. Using the div ergence the- orem, conv ert it to a surface integral o ver the Gaussian surface. Ev aluate the in tegral to ob- tain the (radial) electric \ufb01eld at r . Y our result should con tain a term prop ortional to the c harge q on the inner sphere, and another small term prop ortional to m 2 0 . Integrate this electric \ufb01eld from R 1 to R 2 to solv e for the p oten tial dif- ference v that would b e measured b etw een the inner and outer spheres. Assuming that q =0 , R 1 = 0.5 m, and R 2 = 1.5 m, and that V 0 is 10 kV, \ufb01nd the volt- age v b et w een the inner and outer spheres that w ould b e observ ed if the photon had a rest mass m 0 =1 0 \u2212 15 eV/ c 2 .

Griffiths Problems Solutions

Eletromagnetismo

UFJF

Eletromagnetismo

Continue navegando

Materiais relacionados

Perguntas relacionadas

What are the main social problems and their possible solutions according to the text? a. The text does not provide information on social problems...

Regarding the meaning of scientific methodology, mark the correct alternative. a) Scientific methodology helps the researcher to focus on only on...

What are the solutions found for agency costs? a) Market forces, administrative problems, agency costs. b) Administrative problems, market forces...

16 - Considering texts I and II, it is correct to say that in both cases a) people notice the impact of climate change in their lives. b) three cli...

Outros materiais