test1soln - '1 5.0 '2'. (l) |.U‘.'--'o (a) I u (b) I...

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Unformatted text preview: '1 5.0 '2'. (l) |.U‘.'--'o (a) I u (b) I n". (c) I on... (d) l w. (e) I .. (b) I r: ‘3'» (c) ECEZZIS TEST No.1 PROF. WONG Famil Name 6:30-8:00 .m., Manda , Februa 5, 2007 WILLY INSTRUCTOR Given Name Student Number Please Insert Your Name in Full and Student Number Consider the following three points in a sphericai coordinate system (r, 6, a): P = (51mm, 79°, 25°) P1’ = (50mm, 57", 60°) I P2’ = (68mm, 80", 48") Find the position vectors r, rl’ and r2’ of the above three points, respectively, in the spherical coordinate system; the position vectors r, rl’ and r2° of the above three points, respectively, in the Cartesian coordinate system; the distance vectors R] = r — n’ and R2 = r — r;’ in the Cartesian coordinate system; the electric field E at point P in the Cartesian coordinate system due to point charge Q. +14 110 located at point P.’ and point charge Q2 = —14 nC located at point P2’. the electric field E at point P in the spherical coordinate system due to point charge Q1 = +14 nC located at point P1' and point charge Q; = —14 ac located at point P2’. SOLUTION OF PROBLEM (1) Positions vectors in the spherical coordinate system: P =(51mrn, 79°, 25“) :> r = rfir= 51 5, [mm] P1’=(50 mm, 57°, 60°) => r,‘ = ra,= 50 5, [mm] P; = (68 mm, 80°, —18°) a r; = n.8,: 6815,[mm] Position vectors in the Cartesian coordinate system: P =[Slsin(79°)cos( 25°), 515in(?9°)sin( 25°), 51005(79°)]=(45.37, 21.16, 9.73) r = xii,L +yay +zéz= 45.375x +21.16§i3r + 9.73132 [mm] P.‘ =[505in(57°)cos( 60°), SOsin(57°)sin( 60°), 50cos(57°)]=(20.97, 36.32, 27.23) n’ == xfix +yéy +28; 20.975:x 4-36.325, -l-2'i".23£i:a [mm] [’2' : [685in(80°)cos(—18°). 685in(80°)sin(—18°), 68008(80°)] = (63.69,—20.69, 1 1.81) r;’ = xii" + yéiy + zfiz= 63.695It —20.69€t,r +11.81§iin [111111] Distance vectors in the Cartesian coordinate system: R; = r — r1’ = +45388x +21.16fiy + 9.7382 -2o.97a, 46.3st 47.2352 = 4-24.405! 45.165, —17.50§z[mm] :> |R1| = 33.64 [min] R2 = r — r2” = +45.Z’u7éi,t +21.165y + 933$Z —6'_’i.69€1K +20.69fiy ~1 1.815.z = 48.325, +41.85&, — 2.085: [mm] :> |R2| = 45.731.11.11] ECE22lS TEST No.1 6:30-8:00 .m.,Monda ,Februa 5,2007 PROF. WONG WILLY INSTRUCTOR Family Name Given Name Student Number Please Insert Your Name in Full and Student Number | n--. ((1) Electric field E at P in the Cartesian coordinate system due to Q] at P1” and Q2 at P2“: Q| (r_r1l) Q2 (r_ré) E: '3 ' 3 4m:" |r—r,] 41130 lr—r2| + _ (+14x10-°)[C] (+£24.40éix 45.165), —17.50&z)[mm] 41t(8.854x10"3[Ffm]) (33.64[mm])3 (—14x10‘°)[c] (—18.322'ix +41.85é_¥ —- 2.085z)[mm] 4n(8.854x10_'2[F;’m]) (4S.73[mm])3 (14,000)[N.m3] +24.40&x 45.1655, 47.5031: 48.3253” +4l.85z'iy — 2.08:3z 41:(8.854}[C] 38,068.69254x10‘6[m2] 95,632.08052x10'6 [m1] 3,305.298547(+24.40§1x 45.165), 47.5051) —1,315.7‘55063(—l8.325x +41.855y — 2.0851) [NICI = +80,649.284555x ~50,108.32597éy -57,842.72457éiz +24,104.63275éx —55,064.34939§Y +2,736.77053 152 [NIC] 104,3’5391'3'321x —105,173213716451)r -55,105.95404-r?1It [NIC] ill 104,7s4ax 4053735, 45,1065: [Wm] => |E1'=' 1ss,340 [Vim] l n (6) Electric field E at P in the spherical coordinate system due to Q. at P.’ and Q2 at Pz’: E I" 104554591x —105,1735y -55,l 06551z [Vim] [+104,7S4sin (79°)cos{25°) —105,173sin[79°}sin(25°) —55,106cos(79°)] a, + [+104,754cos(79°)cos(25°) —105,1?3cos(79°)sin(25°) +55,106sin(79°)]aa + [—104,754sin (25°) —105,l73cos(25°)]5¢ Ill [93,195 —43,631 4051515, + [13,115 —3,431 +54,094] at, + [—44271 495319154, 39,0493r + 63,728 as — 1321,5905, [Wm] :> [E|5158,340[me] ECEZZIS TEST No.1 6:30-8:00 .m., Manda , Februa S, 2007 PROF. WONG WILLY INSTRUCTOR Famil Name Given Name Student Number Please Insert Your Name in Full and Student Number 5. o v. (2) Consider the circular surface charge distribution of inner radius a, outer radius b, lying in 930 Sin the xy plane and having a non-uniform surface charge density of p5 (find) = , as shown below: 2 4 [1% (a) For the above charge distribution, find an expression for the electric field E(z) at point P. Note that pso is a positive constant, and be sure to justify any symmetry arguments, if used. You may find some of the following integrals and trigouometric relationships useful: udu _ —1 du _ u “d _e‘”‘(au—l) a sin(26) = 25in6cost9 sin2 6 = %[l — cos(26)] cos2 6 + sin2 6' =1 1.0% (b)1f a = 0.5m, b = 1.0111, pso = 35.416x10_12C/m, and a single electron of charge e = —1.602><10"9C is placed at the center of the charge distribution shown above, i.e. at the origin 0(0,0,0) of the coordinate system, find the magnitude and direction of the electrostatic Coulomb force F.= experienced by this electron. ECEZZIS TEST No.1 6:30—8:00 .m.,Monda ,Februa 5, 2007 PROF. WONG WILLY INSTRUCTOR Famil Name Given Name Student Number Please Insert Your Name in Full and Student Number SOLUTION OF PROBLEM (2) “I. ' ' . I: I : r r I - l = 4. u . (a) Posmon vectors. 1' p 39 p cos¢ a“ + p sm¢ a)r and r z a, Distance vector: R = r — r'= -p'cos ¢'ax — p'sin Way + z a! Magnitude: ]R| = |r —~ r'[ : 1/(—p‘)2icos2 ¢'+ sin2 ¢' i+ zz = 1/(p‘)2 + 22 Point charge: d0 = psds = ps(p'd¢'dp‘) = [Eggwdp'dfl dQ = (Pso Sin ¢'Xdpld¢') = Pso Sin ¢ld¢'dp' dQ i 47:80 |R|3 pan Sin ¢' d¢' dp' dE = —-—_72— 4x60[(p')2 + 22? _ ‘d ' I ‘ I I dE = 438 :22 R cosgfi 5mg} d¢ ai O -p p'dp' - . . 5" sm2 gr} d9} 3’ +4 '2 :14"? Mul(p)+zl psozdp' . . . —~—— 5111 d a 47rza'fl[(,4o')2 + 22F: ‘6 ¢ ____ pSI] _p'dpl _l_ - l t dE_[4zso][[(p') +22T’2][2sm(2¢ was!" +[4:: [é-[l—cosa‘fi'fldgfi'] a, a 2 Electric field: dE = (—p'cosfial —p'sin¢'ay + 2 at) + [(p'f +221” +{ M 47:80 zdp' - . = p50 "‘p'dp' I _ - u c Integratmg. E [47:80] a I[ Sln(2¢ a, 2 1 firm ".__-p'dp' 1_ . . ECEZZIS TEST No.1 6:30-8:00 .m.,Monda , Februa 5,2007 PROF. WONG WILLY INSTRUCTOR Famil Name Given Name Student Number Please Insert Your Name in Full and Student Number b d I 2x - {41:0} iw—kflfizzffl] Jew-am, The sine and cosine functions integrate to zero between 0 and 27:, yielding _ pan . [was]Ml!“ p50 " -p'dp' #930 b p'dp' E: — =_ — [47:80] a [(p')2 +z2rfl](fl) a” 450 a [(p')2 +2217”£2 a’ l p'-—-b a : pm 3' 46-0 l(pl)2 +22 p'=a fl'=h a? = _ p50 _ 1 480 V(p')2 +22 ' p =3 Therefore, E(z)=fi 4—1 «(—1 [V] [N] —-—-—--— a — or — 48a[ b2+z2 a2+z’] ’ m C .0». (b) Given: a=0.5[m], b=1.0 [[11], p50 =35.416x10_12[Cl’m], e=~l.602x10_19 [C] At the origin 0(0,0,0) of the coordinate system #0501”; =1_J_ = _ -_ E 12(0)_4£0[b Jay [I 0.5]:1’ (1 2)a,_ sic] Force experienced by this electron is given by: Fe = q E :(e) (~ay) : (—1.602x10"°)(—ay) [N] Therefore, FI = 1.602 X 10ng a" [N], which is pointing in the +y direction ECEZZIS TEST No.1 6:30-8:00 .m. Manda , Februa 5,2007 PROF. WONG WILLY INSTRUCTOR Famil Name Given Name Student Number Please Insert Your Name in Full and Student Number 5. u "x. (3) A wire, which is being used to charge a capacitor as shown below, lies a10ng the z—axis. Wire Parameters Radius = 5 m p Length = l m It is known that within this wire, the charge density is given by MW) = pvoe'tfi - e'”) tC/mfi where ,0,0 = 1 uCr'm3, y= 200 m—I and a = 3500 s". 2. (a) Make use of the differential form of the continuity equation to determine the current density, J(p,z,t), within this wire. You may assume that J only has a 2 component and that at I = 0 s and z x 0 m, the current density has a value of zero. :. 5% (h) After a long time has passed, the capacitor becomes fully charged. For this steady-state condition, make use of Gauss's law to find the magnitude and direction of the electric field E at z = 0.5m and p = 10mm. Hint: You may assume that the wire is infinitely long, since its radius (5mm) is small compared to its length (1m) and the 10mm distance from the wire axis (i.e. the z axis) is also much smaller than the total wire length (1m). The following integral may be of use to you: J'ueWu = (u — 1)e" +K SOLUTION OF PROBLEM (3) 2. 52,. (3) Continuity equation: div J = —§§-"- where J = Jz 2, f dzv J = —%€‘- z -% we” (1 e'“ )1: — assumeflweflll = — argue—’9‘“ a]- _ a, dzv J = = —apwe ”* ECEZZIS TEST No.1 6:30-8:00 .m. Manda , Februa 5,2007 PROF. WONG WILLY INSTRUCTOR Famil Name Given Name Student Number Please Insert Your Name in Full and Student Number sz’zst) = _ apvfle‘mum Idz = — “Roe-’9‘": + K Jipaoso) = - apwe‘mm + K = 0 => K = 0 Therefore, J = J: (p, 2,011; = ~apwe'wz ax 1 (b) Under steady-state conditions: pv(p,t) = pr( p, 00) = pwe'am [Cfm3] For a cylindrical Gussian surface Gauss's law gives: 80 SHE - ds = Qm,_ L Ear I. Saw 2!: an O; 15,, sag-mm; I I [pie-wwde L Ear L 2x5mm goEppI : pvl] I]. Je-WMMM 1 p‘=0.005 80510;)sz = pvflL27: —2 I(_ypu )e-m'd(_ypl) :3 _ +K y p'=0 =0_005 E — pr“ [(-—;«p'—1)e*“="+K];o .. “'0 [(—o.005y—1)e‘°-°°”+1] “’ _ risop rise/9 pl.0 —2 10*[C/m3] 2 E»: 2 _+1 3 2 2 —12 2 2 la? 9/ 80p 3 200 [llm ]8.854x10 [C le ]0.01[m] e Ep = [1— 3]=(282.3532562X0264241117) [5 2 x8.854x10 [C] e c E3, = 74.61 [X] m Therefore, E=Epap= 74.61% [X] m ...__4 «urn ...
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test1soln - '1 5.0 '2'. (l) |.U‘.'--'o (a) I u (b) I...

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