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411-exam1.sample2.soln

411-exam1.sample2.soln - Physics 41 1 Exam I V D(r,t =...

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Unformatted text preview: Physics 41 1 Exam I March 13, 2001 V . D(r,t) = 47rk1 6,,po(l', t) V . B(r, t) = 0 V x E(r, t) = —k36La:”)— _ h a 6D(ra t) V x H(r,t) — k1 #060 —6t V . J(r,t)+%§’—Q—= 0; 5(g(x))= 2‘. 5|: I- X.) i E X=XI df= dr 0 Vflr) system D(r )= sE(r) + 47l'k2fiJ003t) F = q[E +k3v x B]; 5(k—k')5(a)—w')= 1 D(r)= eE(r). esu (Electrostatic) Name—ki'é/_— 1. (50 points) 2. (25 points) 3. (25 points) total ._____(/ 100 points) [D1(r, t) — D2(r,t)] . n = a(r,t) [B1(r,t) — B2(r,t)] . n = 0 n x(E1(r,t) —E2(r,t)) = 0 n x (H1(r,t) — H2(r,t)) = 2(r,z) (I>(r)=z,—;g: ”VJ FLY—9;! d3x + (1),, ”H e—ilr-(k—k’ >'(W")‘1d3rdt all space, time (27!)4 m f(r')5<3>(r —r’)dx'dy'dz'= f(r) contains r emu (Electromag) Gaussian Heaviside—Lorentz leO‘?&V 1:13.014" SI (MKSA) 60 = :3; 8.854 187 10-12 Farad/m (SI) ,a0 = 47:10'7 1.256 637 10“6 Henry/m 5+ r376.730 ohms ((2) 376.730 0105/c2 h 1.0546 10‘34Js 6.582 10’léeVs c 1.602 10‘19C0ulomb 4.80 1016statcoul (esu, Gaussian) Mich (S1) = 13.104 z—Z (esu, Gaussian) ’1 Bohr radius, rH 0.529177 A 0.529177 10"0 m (SI) =J 13.605 eV e2 4160(2ry) _| J {—H (esu, Gaussian) D(r)=-— —1—” ”(’0)“ -—r—l3)d3r'+Do(r),SIunits VoDo(r)=O I)|r— r| ¢(r) = H]! [resurface fi<r>~<s1E1(r) —e2E2(r)>= cm; ‘2—‘3121 = ¢(x,y,z) 12(2) = 3.0—0 Ir r thdS' + ¢o. SI units p(r)= 0(r)6(F(r))|VF(r)| 3(r)'(81E1(1’) —82E2(r))= 0; 3:21 0(r,5)5= 7mm; ¢+(r) — ¢—(r) = -%fld(r) ¢(r) = —— “H” (r)—— —l) -dS’ = —— HI Mada Irr- r 'l Wm = —%p(r) ¢<r> = % m p(r’)g(r,r')d3r' V2g(r,r’) = 5(3)(r-r’); g(r,r0) = 44:11.0! +F(r,ro) ¢<ro)= m g(r,ra)p<r)/sd3ri If [¢(r)Vg(r,ra)—g<r,ro)V¢<r>1-fi<r)ds volume bounding surface ¢D(r0) = % m GD(r,ro)p(r)d3r + H ¢(rs)VGD(rs,r0)odS V0110“ bounding surface VZGN(r,r0) = 6(3)(r—— r0); fl VGN(r,ro) . fi(r)dS=1 bounding surface ¢(1‘0)= é— m G~<nro>p<r>d3r +¢avg+ If G~<r,ro>V¢<r)-fi<r)ds V0110“ bounding .s‘urface [VGN(r,ro)]-fi(r)= VFN(r, r0)+V I aim: 1 47r|r— fol mm = fl ¢<rs)L_DVGD<rs,ro)-ds; ¢L-~(r)=— [I G~<rs,ro)v¢L.N(rs)-ds bounding bounding surface .s‘urfuce S(l‘,t) = E(r,t) X H(l',t), Welectron = 92/251 Wi = % J‘J‘J. lv¢i(r)‘2d3r volume N W1; = 8 m V¢i(r) - V¢j<r>d3h QFFZICW" C“:— %I , V}: 0 volume W=% ' ini=%zc.,v,.vj Imvm = jflF<w<r>W<r>>d3r _ —_1_ 1 —R/r —F¢/"2_ a_x,- flaw/6x.) 2 ”Wm z 0 GD(”) 7[ |r— (Rz/r’2)r |] a lr- rl 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. h is the surface normal Each part is 4 points. Note that all these questions can be answered using the formulas on the cover sheets. “1.52. . 35;: VuSW—t‘. a) If p(r)— _ [Q/a3] then ——I”V0[p(r)vl_— 0|]d3— a, a3 wee 11‘ ml —l l 4/: |r—ro| for the solution to electrostatic potentiajl d§e_ to a oint charge +5Q, located at —z I ¢D(ro) = ‘iii Mfr + “W <th \7" —— -_~ odS volume .v wface R— —7 .0 c) A uniform surface charge density, 0, is placed on each of the following non conducting surfacesln each case give the correct expression for the volume charge density, ,0(r) i) a sphere centered at the origin with radius- — R [g (0 6" 3 (r- R? ‘ l W V- = V e R 7- 0 ii) a cylinder with radius b and length L>>b, centered at the ori in \"v i- ‘ ' (ya—welt ‘kl MC- (ab-- olvF=l iii) a thin disk with radius R, 1n the yz plane. 6‘ Sfx) Hoff , w) at) e, (a?) iV)x2+yT+%=1 cs 10+"?qu [ d)For what kind of interface surface charge density is ii . E(r), continuous? 44"}:le “an“ C\”"”"tQ‘ \ “2““ ‘ e) “Tall space‘:—— If Qmml (Lg:|§(I(X,yaz))inid3x = Q +°4 “ /(x ,,y z)=0d f)The total energy stored in the electric field of an electron is e2/20 , where a is the radius of a shell upon L = m Cl which all the electric charge is assumed to reside. When this energy is set equal to the rest mass energy for 2: __ .31. the electron, mecz, one finds an effective charge radius for the electron— — 1.4xlO ‘3 cm What IS the e zmc approximate charge radius of the proton which 1s about 2000 times more massive? 1 V l0 n c m a 0.; g) The capacitance of two nested conductors lS given by [3(1 — b/c)] 1.1n Gaussian units 9 = What 15 the capacitance in S. 1. units? , 4E £41". l: I t - ”A! I I if [3 (1—5/ )1 ‘05 A Q3 — -(1) qx loner"; . 7.; h) If V . D(r) =Q/02§(r— a)é‘(cos¢9— l)0‘(¢— 7r/2) then D(r)= 511 g2 13:33 P 01km; ro- , q-E- :0 1 “5‘3 X : qs‘me £11420 ;:\{ ASH/1054:1141 -o i) Which of Maxwell’s equations gives rise to Gauss’ law? _. En) = £ ? j )The (normal or tangential) id 1mg en‘tw'tl component of the electric field is continuous at a conductor—air interface? k)Iff[0(r)]= ln[b/0'] then the variation §afl0(r)] = \ L0 8 r I) In Gaussian units, what is the electrostatic potential of a charge Q, placed at (x,y,z)=(0,4,0)?. Assume the (NF) 2 E2 A POtential is zero at infinity, ____l:;j_~/_l (<5) c=€~=> C: bomb/711 NOTE: In this examination MathCad ready integral means an integrand with no delta functions and the proper limits for the integral. Substitute all numerical values and carry out all operations before substituting functions into the integrand. But do not integrate over anything other than delta functions. 2 An isolated, charged surface is defined, in cylindrical coordinates (p, ((2,2), by the relationships 2 — 5 p = 0 1cm < z < 2cm. The surface charge density is - 2 G(p,(p)= ZETZ (I; with q = 7 X 10-6 Coulombs (a)(15pts )What is the total charge on the surface? (A MathCad ready integral lS sufficient) Q- 3“ [919W :8“ at?) 8mm; (mm) Pulp 3;, d2' (5) VendoSeS Come sari-ace. (lOpts, 2b) Taking the potential to be zero at infinity give the integral expression (MathCad ready) for the electric potential at p = 20 cm,(p = 7r/2, z = 40 cm I . . — \ ‘ I NJ’kf You Cannot Swhp‘a uge cvimalflcai coordinaks Since, CohBSLX'FDUCt\S not mct‘hmdeh Wt ls, (l: \nag a, swindle" Soviluce, area, “nano- (filmltr 52,9. duaqmm ‘ +£7MWl€l AS = AX Ax \lm'ctsarede L J PG.) 33 . Cram 4h: “Shadow” _ ' ' SSS ‘T'ZT' ’“ ilflfi n\ opme Surceee on (M r) = “‘50 “A _\~\ [36‘ .J Waprojecho'} 9““ V encloses C0“!— L'PC’ \ horwa kprojec‘ho'n / i H 3‘ cl ’ale' "5 pm¢~ ' sin4> j/ifli—iflv (3 9 ‘P : 7—” 8381‘” H‘ ex put-l x:20.cos4>=o ‘I: '20 Sum? :20 2:140 O b Scwuhon polflt ' 1: i Source/chums powt .I 3. The energy stored in the electrostatic potential between two nested conductors is given by 400a — 02 + 0.3] W = % III [ 20 d3x = % III |V¢inner conductor|2d3x a where the total volume, V: 85a3, and the outer conductor is assumed to be at zero potential. (a) (15 points) Find the 0‘0 which minimizes the electrostatic energy §W - 20.18% [40. ~2Vo—136‘93X =0 Vali We. Mt “Mt W‘PV’; 0 _ L ‘ 3 1+ _16~] S 6’0 '1'— O . _ 2a,, -’%b’a-W:I"Cf 0 86C“: C) 7? no Field 6; : 20. L :7 "0 du‘Hoslahc pokwha\ Tim‘s is a, ‘Muml case, (b)(10 points) If the potential 18 zero on the outer conductor and aToépn the innermost conductor, what is the capacitance of this system? 1. 35$ Lmah Lia) ”Hr—+03] VJ- 3w“ ...
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