20091ee1_1_hw2-sol - as The cylindrical surface p = s cm...

Unformatted text preview: as. The cylindrical surface p = s cm contains the surface charge density, p. = 5e'ml'l nC/m’. a) What is the total amount of charge present? We integrate over the surface to find: Q=2f£k&m(M)#hnC=%(.M)(%) e-m‘l: =11an h) Howrnuchfiuxleavesthesurfaeep=8cnn lcrn<z <5crn,%°<¢<m°? Wejust integrate the drarge density on that surface to find the flux that leaves it. D" 90' 90 _ 30 —1 '°° e = Q’ = [01 A. 5e-”'(.08)d¢de nC = (W) 2«(5)(.03) (W) 9',” .01 = 945 x10"nC = 9.45pC 3.8. Use Gaus’s law in inteyal form to show that an inverse distance field inquheriul coordinates, D = Aa,/r, where A is aconstant, requires every spherical shell of 1 In thitkneu to contain AKA ooulomhs of charge. Dos this indicate a continuous charge d'strihution? Ifso, find the charge dendty variation with r. Thenet outward fiuxofthiefieldthroughasphericalmrfaceofradiusris " A 6=fD-dS=/h/ 7e.-e.r’ein0d04¢=4mr=q..d 0 0 Weeee fromthisthatwitheveryincreueinr by one In, theencloseddlarge increases by but (done). It '3 evident that the charge demity '3 continuous, and we can find the density indirectly by constructing the integral for the enclosed charge, in which we already found the latter from Gaus’s luv: Q“:=41rAr=[ck/O'A’flr'Hi-Q’sinfldr'wdé=41r/o"p(r')(r')2dr' 'Ib obtain the correct enclosed charge, the integrand mmt he p(r) = Ali-9. 1.14. Show that the vector fieltb A = a, (sin20)/r’+2no (sin9)/r9e.nd B = rcosOer-a. mmrywhere penile] to end: other: Using the definition of the cross product, we find sin29 _ 2sin0cosO r r AxB=( )a,=0=|A||B|sinon Identify n=q, endsosin0=0, and therefore0=0 (they’re parallel). 1.18. Transform the vector field H = (A/p) 3‘, where A is a constant, from cylindr'wd coordinates to spherical coordinates: Fiat, the unit vector does not disuse, since a; '5 common to both coordinate systems. We only need to exprea the cylindrical radium, p, as p = ran, obtaining A "("9) = rsinO 8‘ 1.27. Themrfaeeer=2end4,0=30° end50°,end¢=20°e.nd60° identifyecleeedaurfeee. a) Findtheencloeedvolume: Th'nwillbe 0. Vol=/ I fr’anowa=w ' 30° 9 when degrees have been converted to radians. b) Find the total me of the eneloa'ng surhee: Area=/;v1:0.(49+29)sin0&d¢+/:L..r(sin30°+ein50°)drd¢ +2/m..‘/:rdrd0=fl c) Findthetotellengthofthetwelveedgmofthemfeeez Length=4fdr + 2’ (4+2)d0+/ (43in50°+43in30°+2ain50°+2sin30°)d¢ 2 30° ao- =1ua d) Find the length ofthe longest stru'dnt line that lies entirely within the surface: Thiswill be fromA(r=2,0=50°,¢=20°) toB(r=4,0=30°,¢=60°) or A(.c = 28in50°ooe20°,y = 23in50° sin20°,s = 200350') to B(.c = 49inW°00360°w = 48in30°dn60°,z = 400330°) or finally 41.44.052.120) to B(1.00,1.73,3.46). Thus B _ A = (—0.44, 1.21, 2.18) and mm=|B—A|=M 2.3. Point charges of EOnC each are located at A(l,0,0), B(—1,0,0), C(0,l,0), and D(0,—1,0) in free space. Find the total force on the dune at A. The force will be: 5‘: (50x 10"?[347 +_ 11—0,. + R34 _] “to IRCAI' IRIMIa IRBAP when RCA =a. —ag, R04 =ag+ag, and R34 =24». The magnitudes are |R¢u| = |RDA| = J5, and mm = 2. Substituting these leach to _(50 x 10-9)2 [ 4m 2—f+m+8]ag=2l.lw when distances are in motels. 2.11. A charge 00 located at the origin in free space produces a. field for which E, = l kV/m at point ”-2! 1,—1)' a) FinonzThofieldathillbe Since the 2 component is of value 1 kV/xn, we find 00 = -4ir6061" x 103 = —l.63 g. b) Find E at M(1,6,5) in oarteaian coordinates: This field will be: —1.63x 10-“ a.+6_a,+5a,] Eu:— mo [1 + 36 + 2511.. or Eu = 40.11... —180.63a, _ 15053.... c) Find E at M(1,6,5) in cylindrical coordinate: At M, p = J1 +36 = 6.08, ¢ = tan"(6/l) = 80.54°, and: = 5. Now 13, = By .a, = —ao.11 eoa¢ — 18053.51. 4. = 483.12 E, = By . a, = —30.11(—s'n¢)—180.63ooa¢ = o (as expected) so that Eu = -183.122 — 150.530.. d) Find E at M(l,6,5) in galleried coordinates: At M, r = Vi +38+25 = 7.87, ¢ = 80.54° (as before), and 0 = 0034(5/7487) = 5058'. Now, since the charge '3 at the origin, we expect to obtain only a. radial component of Eu. This will be: E, = Eu -a,. = —30.lla'n9006¢—180.633in03in¢—150.53o090 = —237.1 2.16. Within a. region of free space, charge density '3 givon as p, = par/a C/ma, when p0 and a are constants. F'indthetotal dumlyingwithin: a) the sphere,r5¢: Thiswillbo Q.=['/:[¥r99modrdod¢=u£$¢=fl b) theoono,r5¢,050$0.lr. & (Mir 3 _ fl . = a — . = . ’ Q._/o [o [0. “#modrdoau 2r 4 [1 ”(01m comm c) thoregion,r5a,05050.1t,05¢50.21r. 0.2! 0.1! 00? . a 03* a 0.: [o [o I: Tfimodrdoaauuma (y) =0.0024irm ...
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