HW#5_Solutions

# HW#5_Solutions - Problem 4.11 pi, = 0; m, = P4“: = :tP...

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Unformatted text preview: Problem 4.11 pi, = 0; m, = P4“: = :tP (plus sign at one end—the one P points toward; minus sign at the other—the one P points away from). (i) L >> (1. Then the ends look like point charges, and the whole thing is like a physical dipole, of length L and charge Pn’ur". See Fig. (a). (ii) L < a. Then it‘s like a circular parallel-plate capacitor. Field is nearly uniform inside; nonuniform “fringing ﬁeld" at the edges. See Fig. (b). (iii) L z a. See Fig. (c). a 'Illlllllllll v (a) Like a dipole (b) Like a parallel-plate capacitor Problem 4.13 Think of it as two cylinders of opposite uniform charge density :tp. Inside, the ﬁeld at a distance 3 from the axis of a uniformly charge cylinder is given by Gauss‘s law: E21rs€ = 5mm => E = (p/Zco)s. For two such cylinders, one plus and one minus, the net ﬁeld (inside) is E 2 E4, + E- = (p/250) (5+ — 5.). But s+ — s- = -d, so E = —pd/(2co), where d is the vector from the negative axis to positive axis. In this case the total dipole moment of a chunk of length l is P (nazl) = (puff) d. So pd = P, and E = -P/(260), for s<a. Outside, Gauss’s law gives E21rs£ = fame”! => E = is, for one cylinder. For the combination, E = 2:08 E++E_=%(§—f—§—;),where s - 2. 2i: " 54:21 :2- 9 2i ._1al «.1 s-d'zl d ml 81 — (BZF2)(8+4:FS d) _32 sq:2 14:6—2 -8—2 54:5 Its—2 1 (s-d) d l _2 3—2 (s i s 62 q: 2) (keeping only lst order terms in d). (s+s(ss'2d) — — (s — s(:2d) + = :12- (2s(ss'2d) — d) . A com: +i+ l mm» .l. V II ca for s > a. Problem 4.19 With no dielectric, Co = Aeo/d (Eq. 2.54). In conﬁguration (a), with +0 on upper plate, —a on lower, D = a between the plates. E = 0/60 (in air) and E = 0/6 (in dielectric). So V = 5% + 5% = 2‘ 4 (1+ in). 0 e Ca 26,. 0,, = g = 431A (1H2 cr) =, _C‘,) = “Fer. In conﬁguration (b), with potential difference V: E = V/d, so a = (057 = eoV/d (in air). P: 6(1er = coer/d (in dielectric). so ab = —eoer/d (at top surface of dielectric). mm = eoV/d = a; + m, = a; — cox.V/d, so a; = eoV(1 + x.)/d = eoe,V/d (on top plate above dielectric). =}C—Q—.l. aﬁ+a£ —i K+€K ——ﬂ 1+6' 9—6—14.” b_V_V 2 ’2 ‘2v ‘°d °d" 'd 2 '00- 2 ' - - c c _ 1 2 _ 1+er °-4e. _ 1+2er+4<’-4¢. _ 1—e, ’ [Which is greater? 5; — 5; _ ﬁg: — _ i—QWLT _ —r—,(l+¢r, _ W > 0. So 0., > 0a.] Ifthe 1: axis points down: —l-_-_- a» (top surface) I a (top plate) I 2 - X V (—5? Problem 4.21 Let Q be the charge on a length I of the inner conductor. _ _ _i. _ Q _ Q fD-da — D27rs€—Q=>D—21rs[, E—21r608[(a<s<b), E—2ﬂsl(b<r<c) a b _ . _ Q d_s ‘ .2. e- o 2 e_o g V - _/c E (II—A (2160!) s +,/,, (27rd) s -21rcot [1n(a)+ e 1n(b)l' C __ Q _ ha; 7 ‘ v7‘ ln(b/a)+[1/e,.)ln(c/b)' Problem 4.24 Potentials: Vamp-,0) = —Eor c039 + Z %[H(0059), (r > b); vm.d(r,a) = 2; (AH + 3‘25) mcosa), (a. < r < b); Vin(r,0) = 0. (r < a). Boundary Conditions: Vout = Vmed’ (7 = b), (ii) 68—39;,“ = e 8V, , (r = b), (iii) Vmed = 0, (r = a) (i) => —Eobc059 + X: %H(cos9) = Z (Aib‘ + 1%)Mcos0); (ii) => 5,: [ow-l — (1+1)% P;(cos€) = -Eo cosO-ZU + 1)%H(0089); B _ (iii) 5 All-'4‘ + (ti—:1 = 0 => B: = -a2'+1Ag. Forl¢lz B 2144A (i) b?!) = (Alb! — ab,“ ‘) :9 Bl = Al (1,9144 _ an“); .. _ amid/1 Bl z 2! 2! I _ _ (ll) €r|:lA1bl 1+(l+1) “+2 I] =—(I+I)W => B[=-€1-A1 “+0 + => Al—Bl—O. For! = 1: 3 (l) -Eob+%l'=A1b-abﬁl => Bl-Eob3=A12(ba-03); 03A; El 3 3 3 (n) 67(1114-2 b3 )=—Eo—2F => —2B;—Eob =erA1(b +2a). _3130 So -3Eob3 = A1 (b3 — as) +£r (b3 + 203)] ; A] = ~3Eo 23 V“"(”9) _ WW (’_r“)c°so’ E(r,9) = -VVmed= ...
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## HW#5_Solutions - Problem 4.11 pi, = 0; m, = P4“: = :tP...

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