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Griffith...ual-pdf) - introduction to electrodynamics (solutions) - ch04, Exercícios de Física

exercícios resolvidos, do livro do Griffiths de eletromagnetismo, separados pos capítulos.

Tipologia: Exercícios

2010

Compartilhado em 26/08/2010

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Baixe Griffith...ual-pdf) - introduction to electrodynamics (solutions) - ch04 e outras Exercícios em PDF para Física, somente na Docsity! Chapter 4 Electrostatic Fields in Matter Problem 4.1 E =V/x = 500/10-3 = 5x 105. Table 4.1: a/47r€0 = 0.66x 10-30, so a = 47r(8.85x 10-12)(0.66x 10-30) = 7.34X10-41. p =aE =ed ~ d =aE/e = (7.34x 10-41)(5 x 105)/(1.6 x 10-19) =2.29 X 10-16 m. d/R =(2.29 x 10-16)/(0.5 x 10-10) = 14.6x 10-6.1To ionize, say d = R. Then R = aE/e = aV/ex ~ V = Rex/a.= (0.5 x 10-10)(1.6 x 10-19)(10-3)/(7.34 x 10-41) =1108v.1 Problem 4.2 First find the field, at radius r, using Gauss' law: JE.da = E~Qenc,or E = 4;<0~Qenc. l r 47rql r - 4q [ a - ( a2)]l r Qenc = pdT = - e-2r/ar2dr = - --e-2r/a r2 + ar +- 0 7ra3 0 a3 2 2 0 2q [ ( a2 ) a2 ] [ ( r r2 )]= - a2 e-2r/a r2 + ar +"2 -"2 = q 1 - e-2r/a 1 + 2~ + 2a2 . [Note:Qenc(r --+ 00) =q.] So the field of the electron cloud is Ee = 4;<0~ [1 - e-2r/a(1+ 2~+ 2~)]. The protonwill be shifted from r =0 to the point d where Ee = E (the external field): 1 q [ 2d/ a ( d ~ )]E=-- 1-e- 1+2-+2-.47r€0d2 a a2 Expandingin powers of (d/a): e-2d/a = 1 - ( 2d )+ ! ( 2d )2 - .!. ( 2d )3 +... = 1 - 2~ + 2 (~)2 - ~(~)3 +...a 2 a 3! a a a 3 a = 1- (1-2~+2(~r -~(~r +..-) (1+2~+2~) d cP. d cP. d3 cP. d3 4~= r - r - 2t - 2:+ + 2t + 4:+ + 4:+ - 2:+ - 4:+ + - - + .. .a d2 a d2 d3 d2 d3 3 a3 4 ( d ) 3 = 3 ~ + higher order terms. ( d d2 )1- e-2d/a 1+ 2- + 2-a a2 73 74 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER 1 q ( 4 d3 ) 1 4 1. I 3 IE = -- -- = --(qd) = -po a = 311"!:oa.471"€0dl- 3 a3 471"€03a3 371"€oa3 [Not so different from the uniform sphere model of Ex. 4.1 (see Eq. 4.2). Note that this result predicts 4;EOa = !a3 = ! (0.5 X 10-10)3 = 0.09 X 10-30 m3, compared with an experimental value (Table 4.1) of 0.66 x 10-30 m3. Ironically the "classical" formula (Eq. 4.2) is slightly closer to the empirical value.] Problem 4.3 per) = Ar. Electricfield (by Gauss'sLaw): §E.da = E (471"r2)= -!oQenc = EloJ;Ar471"r2dr, or E = ~ 471"A r4 = Ar2 . This "internal" field balances the external field E when nucleus is "off-center" an amount471"r €o 4 4€0 d: ad2/4€0 = E ~ d = V4€oE/A. So the induced dipole moment is p = ed = 2ev€0/AVE. Evidently I p is proportional to El/2.1 For Eq. 4.1 to hold in the weak-field limit, E must be proportional to r, for small r, which means that p must go to a constant (not zero) at the origin: I p(O) :/; 0 I (nor infinite). Problem 4.4 r Field of q: ~ ~ f. Induced dipole moment of atom: P = a E =. 0 Q A 1I"EO r q 411"E:r2 r. Field of this dipole, at location of q (0 = 71",inEq. 3.103): E = _4 1 13( 2aq 2) (to the right).7I"€0r 471"€or Force on q due to this field: IF = 2a (-4 q )2 13 I(attractive).7I"€0 r Problem 4.5 Field of PI at P2 (0 = 71"/2in Eq. 3.103): E1 = 4 PI 39 (points down).7I"€or Field of P2 at PI (0 = 71" in Eq. 3.103): E2 = 4 P2 3 (-2f) (points to the right).7I"€or I 2PIP2 I . .Torque on PI: N1 = PI X E2 = -4 3 (pomts mto the page).7I"€or Problem 4.6 (a) Use image dipole as shown in Fig. (a). Redraw, placing Pi at the origin, Fig. (b). E-- P ( . - 471"€0(2z)32cosOf+sinO9); P =pcosOf + psinO9. 10 ~o Pif/ Z 2 N = P X Ei = 471"€:(2Z )3 [(cos 0 f + sin 09) x (2cos0 f + sin09)] p2 [ A A ]= 4r.€0(2z)3 cosOsinO4J + 2sinOcosO(- 4J) p2 sin 0 cos 0 A = 471"€0(2z)3 (-4J) (out of the page). (b) 77 Outside, Gauss's law gives E27r8£ = .1...p7ra2£:::}E = 1!!£.2a2!, for one cylinder. For the combination, E =<0 <0 s E+ + E- = 1!!£.2 a2 (!:t. - iL ), where<0 s+ s- d s:!: = S T -j2 ( d)( 12 ) -1 ( )( ) -1 ( )( )S:i: 2 a- 1 d s.d 1 d s.d- = ST- 8 +-Ts.d ~- ST- IT- ~- ST- 1:f:-8?t 2 4 82 2 82 82 2 82 1 ( (s . d) d ) . . = 82 S :f: S--;2 T "2 (keepmg only 1st order terms in d). (8+ - L ) = ~ [(s+s~ - ~) - (s-s~ + ~)] = ~ (2S(S.d) -d ).s+ 8- 82 82 2 82 2 82 82 a2 1 E(s) = -- [2(P. 8) § - P] 2fO 82 ' for 8 > a. Problem 4.14 Total charge on the dielectric is Qtot = is O"bda + Iv Pbdr = is P . da - Iv V.p dr. But the divergence theorem says is p. da = Iv V.p dr, so Qenc = O. qed Problem 4.15 (a)Pb=-v,p=-~~ (r2~ ) =-~; O"b=P.ii= { +P.~=k/b (atr=b), }r2 or r r2 -P . r = -k/a (at r = a). Gauss's law:::} E = 4:<0Q;~cr. For r < a, Qenc = 0, so IE = 0.1 For r > b, Qenc = 0 (Prob. 4.14), so I E = 0.1 Fora < r < b, Qenc = (~k) (47ra2)+ I: (~) 47rf2dr = -47rka - 47rk(r- a) = -47rkr; so I E = -(k/for) r.1 (b) fD.da = Qfenc = O:::}D = 0 everywhere. D = foE + P = O:::}E = (-l/fo)P, so IE = 0 (for r < a and r > b);j IE = -(k/fOr) r (for a < r < b). I Problem 4.16 (a) Same as Eo minus the field at the center of a sphere with uniform polarization P. The latter (Eq. 4.14) is-P/3fO.SoIE =Eo + ~p.1 D = foE= foEo+ ~P = Do- P + ~P, soI D =Do - ~p.1 (b) Same as Eo minus the field of:f: charges at the two ends of the "needle"-but these are small, and far away,so ! E = Eo.J D = foE =foEo = Do - P, SO I D =Do - P .1 (c) Same as Eo minus the field of a parallel-plate capacitor with upper plate at 0" = P. The latter is -(l/fo)P, soIE = Eo + !op.1 D = foE = foEo + P, so ID = Do.! 78 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER Problem 4.17 P (uniform) E (field of two circular plates) D (same as E outside, but lines continuous, since V.D = 0) Problem 4.18 (a) Apply JD .da = Q/enc to the gaussian surface shown. DA = aA ::} I D = a.1 (Note:D = 0 insidethe metal plate.) This is true in both slabs; D points down. ~ 2+u (b) D = fE ::} E = a/fl in slab 1, E = a/f2 in slab 2. But f = tofT, so fl = 2fo; f2 = ~fO' lEI = a/2fo, I I E2 =2a/3fo.1 (c) P = foXeE, so P = foXed/(fOfr) = (Xe/fr)a; Xe = fr -I::} P = (1- f;l)a. (d) V = E1a + E2a = (aa/6fo)(3 + 4) = 17aa/6fo,! IPl=a/2,IIP2=a/3.1 (e)Pb=O; ab = +P1 at bottom of slab (1) = a/2,ab= -PI at top of slab (1)= -a/2; ab = + P2 at bottom of slab (2) = a /3, ab = -P2 at top of slab (2) = -a /3. (f) In slab 1: { In slab 2: { total surface charge above: a - (a/2) = a/2, } E - ~ ..( total surface charge below: (a/2) - (a/3) + (a/3) - a = -a/2, ==> 1 - 2fO' total surface charge above: a - (a/2) + (a/2) - (a/3) = 2a/3, } E - 2a ..( total surface charge below:(a/3) - a = -2a/3, ==> 2 - 3fO' ]+u -u/2 (!) @ +u/2 -u/3 +u/3 ]-u Problem 4.19 With no dielectric, Co = Afo/d (Eq. 2.54). In configuration (a), with +a on upper plate, -a on lower, D = a between the plates. E = a/ fO(in air) and E = a/ f (in dielectric). So V = {;; ~ + 7~ = 2~~ (1 + ~) . - ~ - ~ ( 2 ) I Ca - 2fr ICa - V - d 1+1/(r ==> o 1 + fr . In configuration (b), with potential difference V: E = V /d, so a = foE = fO V/d (in air). 79 P =EOXeE= EOXeV/d (in dielectric), so O'b= -EOXeV/d (at top surface of dielectric). O"tot=EoV/d=O'f + O'b= O'f- EoXeV/d,so O'f= EoV(l + Xe)/d = EOErV/d(on top plate above dielectric). Q 1 ( A A ) A ( V V ) AEO ( 1 + Er ) I Cb 1 + Er I=? Cb = V = V 0'2" + O'f2" = 2V Eod + EOdEr = d ~ . Co = ~. [Which is greater? ~ - ~ = l+<r.- ~ = {l+<r)2-4<r= 1+2<r+4<~-4<r= (1-<r)2> 0 So C > C ]. Co Co 2 l+<r 2(1+<r) 2(1+<r) 2(1+<r) . b a. If the x axis points down: I ~ E p ] O'b (top surface) ~ O'f (top plate) ID Problem 4.20 ID.da = Qfenc => D41fr2 = p!1fr3 => D = lpr => E = (pr/3E) r, for r < R; D41fr2 = p!1fR3 => D = pR3/3r2=>E = (pR3/3Eor2)r, for r > R. fO pR3 1 1 R P fo pR2 P R2 I pR2 ( 1 )V = - }00 E. dl = 3EO -:;. 00 - 3E }R rdr = 3EO + 3E ""2 = 3EO 1 + 2Er . Problem 4.21 Let Q be the charge on a length £ of the inner conductor. V - Q Q Q D21fs£ = Q => D = _2 0; E = -2 0 (a < s < b), E = -2 0 (b < r < c).1fS~ 1fEOS~ 1fES~ f a I b ( Q ) dS l C ( Q ) dS Q [ ( b ) EO ( C )] - E.dl- - -+ - --- In - +-In - C - a 21fEO£ S b 21fd S - 21fEO£ a E b' Q I 21fEO IV£ = In(b/a) + (l/Er) In(cjb)" f D . da = C £ - Problem 4.22 Same method as Ex. 4.7: solve Laplace's equation for V;n(s, tj)) (s < a) and Vout(s, tj)) (s > a), subject to the boundary conditions x { (i) V;n = Vout at s = a, (ii) E8~n = EO8~~ut at s = a, (Hi)Vout -+ - Eos cos tj) for S » a. Eot y FromProb. 3.23 (invoking boundary condition (Hi)): 00 V;n(s, tj)) = 2::>k(ak cosktj)+ bksinktj)), k=l 00 Vout(s, tj))= -Eoscostj) + L s-k(Ck cosk</> + dk sink</». k=l (a) air < - .2<r - 0 0 Y«r+1) d x «rH) d X «r+l) d (a) dielectric 2 v - x - -«r+l) d X «rH) d <r+l) d X «r+l) d - (b) air Yx - 0 0 7 (left)d d x (b) dielectric Yx Er7x (Er - 1)7 X -(Er - 1)7 Er7 (right)d 82 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER Plug this into (a) and solve for O"b,using €~ = 1 + X~: O"b - -1 qd/€~ ( ' ) O"b ( ' ) -1 qd Xe - ";!Xel+Xe--Xe+Xe,soO"b=- 3 [ ( ,)/] ; 41T (r2 + d2)2 2 41T(r2 + d2)2 1 + Xe + Xe 2 , { -I qd 1 1 qd/€~ } , 1 qd €rX~/€~ Xe 41T (r2 + d2) ~ [1 + (Xe + X~)/2] + 21T(r2 + d2)~ ' so O"b= 41T(r2 + d2) ~ [1 + (Xe + X~)/2]' , -O"b- The total bound surface charge is O"t=O"b + O"~= 417r qd ~ E' l + X (~- +x. , )/ 2 (which vanishes, as it should, when(r2+d2) ~ ", x. X~ = Xe)' The total bound charge is (compare Eq. 4.51): qt = (X~ - Xe)q _ I ( €~ - €r ) q 2€~ [1 + (Xe + X~)/2] - €~ + €r €~' Iand hence V(r) = ~ { q/€~ + qt } I(for z > 0). 41T€0 ";X2 + y2 + (z - d)2 ";x2 + y2 + (z + d)2 q [ €~ - €r ] - ~ Meanwhile, since ~ + qt = 7" 1 + €' + €r - €~ + €r '€r r r 1 [2q/(€~ + €r)] I(for z < 0).V(r) = 41T€0 ";X2 + y2 + (z d)2 \ I I I I \Problem 4.26 From Ex. 4.5: D = { O'Q (r < a) 41Tr2r, (r > a) { 0, Q ~ -r, } , E = 41Ttt ~ ~r,41T€or (r < a) } (a < r < b) . (r > b) 1I 1 Q2 { I lb 1 1 1 100 1 } Q2 { I ( -1 )I b 1 ( -1 )1 00 } w = - D.EdT=--41T - --r2dr+- -dr =- - - +-- 2 2 (41T)2 € a r2 r2 €o b r2 81T € r a €o r b Q2 { 1 ( 1 1 ) 1 } Q2 ( 1 Xe )= 81T€0 (1 + Xe) ~ - b + b = 81T€0(1+ Xe) ~ + b . Problem 4.27 83 Using Eq. 4.55: W = !f JE2 dr. From Ex. 4.2 and Eq. 3.103, E = Wr<R = Wr>R = Wtot = = { -I 3102P z, 3R ~(2cosBf+sinBO),for fa (.£-. )2 ~11'R3 = 211'P2R3.2 3100 3 27 fa fa ( R3 P ) 2 J 1 ( 2 . 2 ) 2 . - - 6" 4cos B+ sm B r smBdrdBd<jJ2 3100 r (R3 p)2 l 1r 1 00 1 11'(R3p)2 8 211' (1 + 3 COS2B)sin BdB 4" dr = -1 fa a R r 9100 11'(R3P)2 (~ )= 411'R3p2.9100 3R3 27100 211'R3 p2 9100 (-COSB-COS3B)I~(-3~3)1: (r < R) } , (r > R) so This is the correct electrostatic energy of the configuration, but it is not the "total work necessary to assemble the system," because it leaves out the mechanical energy involved i~ polarizing the molecules. Using Eq. 4.58: W = ~ JD.E dr. For r <: R, D = foE, so this contribution is the same as before. Forr < R, D = foE + P = -!p + P = jp = -2fOE, so ~D.E = -2!fE2, and this contributionis now(-2) (~~p:~3) = - ~~ R::2, exactly cancelling the exterior term. Conclusion: I Wtot = 0.1 This is not surprising, since the derivation in Sect. 4.4.3 calculates the work done on the free charge, and in this problem there is no free charge in sight. Since this is a nonlinear dielectric, however, the result cannot be interpreted as the "work necessary to assemble the configuration" -the latter would depend entirely on how you assemble it. Problem 4.28 First find the capacitance, as a function of h: Air part: E = -bL ==> V = ..1L In(bJa) } 41rfOS 41rfO' >.. >'" 10 ==> - =-j >..' = ->..= lOr>". OilPart: D = ~ ==>E = 2>" ==>V = 2>"In(bJa) fa 10 fa 41rs 41rfS 41rf ' Q = >..'h + >"(f - h) = fr>"h - >"h+ >"f = >..[(fr - l)h + f) = >"(Xeh+ f), where f is the total height. Q >"(Xeh+ f) (Xeh + f) C = V = 2>"ln(bJa) 411'100= 211'100In(bJa) . Th t d l' .. b E 4 64' F - 1 V 2 dC - 1 V 2 21rfOXe } I V 2 e ne upwar LOrCeIS gIven y q. . . - 2" dh - 2" In(bfa) . h = fOXe . The gravitational force down is F = mg = p11'(b2 - a2)gh. p(b2 - a2)g In(bJa) 84 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER Problem 4.29 8 (a) Eq. 4.5 :::} F2 = (P2 . V) EI =P2~ (Ed; . uy . PI A PI A Eq. 3.1O3:::} EI = ~ () = -- 4 3 z. Therefore4m:or 1rfoY ~y?jr PIP2 [ d ( 1 )] A 3PIP2 A I 3PIP2 A IF2 = _4- -d 3" z = _4 4 Z, or F2 = 4~z (upward).1rfO y Y 1rfoY 1rfor z y To calculate F I, put P2 at the origin, pointing in the z direction; then PI is at -r z, and it points in the -y direction. So FI = (PI' V) E2 = -PI 8:2 1 - - - ; we need E2 as a function of x, y, and z.y x-y-O, z--r . I I E E ll [ 3(P2' r)r ] h A A A A d hFrom q. 3.104: 2 = -3" ? - P , were r = xx + yy + ZZ, P2 = -P2Y, an ence41rfOr r- P2 . r = -P2Y' E2 = ~ [ -3Y(XX + yy + zz) + (x2 +y2 + Z2)y ] = ~ [ -3XYX + (x2 - 2y2 + z2)y - 3YZZ ]41rfO (x2 + y2 + z2)5/2 41rfO (x2 + y2 + z2)5/2 ~ { -~~2Y[-3XYX + (x2- 2y2+ Z2)y - 3yzz] + ~(-3xx - 4yy - 3ZZ) } ; ~~ 2~ ~ ~ -3z z; FI = -PI (~ 3r Z)= - 3PIP2 Z.41rfO r5 41rfO r5 41rfor4 8E2 = 8y 8E2 18y (0,0) = These results are consistent with Newton's third law: FI = -F2. (b) From page 165, N2 = (P2 x EI) + (r x F2). The first term was calculated in Frob. 4.5; the second we get from (a), using r = r y: P2 X EI = PIP2 ( A 41rfor3 -X)j F ( - ) (3PIP2 -) 3PIP2 A I N 2PIP2 Ar x 2 = ry x _4 4 Z =_4 3 x; so 2 =_4 3 X.1rfor 1rfor 1rfor This is equal and opposite to the torque on PI due to P2, with respect to the center of PI (see Frob. 4.5). Problem 4.30 Net force is I to the right I (see diagram). Note that the field lines must bulge to the right, as shown, because E is perpendicular to the surface of each conductor. E 87 (d) Figure (b) works the same way, but Fig. (a) does not: on the flat surface, P is not perpendicular to ft, sowe'd get bound charge on this surface, spoiling the symmetry. Problem 4.37 Eext = ~ 8. Since the sphere is tiny, this is essentially constant, and hence P = €oXe/ Eext (Ex. 4.7).27r€08 1 + 3 ( ) ( ) ( ) ( )( ) 2 ( ) ( )F - €oXe ~ ~ ~ 8dr - €oXe ~ ! -1 8 dr- J 1 + Xe!3 27r€08 d8 27r€08 - 1 + Xe/3 2no 8 82 J -Xe (~ )2-~7rR38 - - (~ ) )..2R3 81 + Xe/3 47r2€0 833 - 3 + Xe 7r€083 .= Problem 4.38 The density of atoms is N = (4/3)7rR3'The macroscopic field E is Eself + Eelse, where Eself is the average fieldover the sphere due to the atom itself. p = o:Eelse =} P = No:Eelse. [Actually,it is the field at the center, not the average over the sphere, that belongs here, but the two are in fact equal, as we found in Prob. 3.41d.] Now 1 p Eself = - 47r€0R3 (Eq. 3.105), so 1 0: ( 0: ) ( NO: )E = - 47r€0 R3 Eelse + Eelse = 1 - 47r€oR3 Eelse = 1 - 3€0 Eelse. So P= No: (1 - N o:/3€0) E = €oXeE, and hence No:/€o Xe = (1 - No:/3€0)' Solving for a: No: No: No: ( Xe )Xe - -3 Xe= - =} - 1+ _3 = Xe,€o €o €o or €o Xe 3€0 Xe 3€0 ( €r - 1 )a = N (1 + Xe!3) = Ii (3 + Xe' But Xe = €r - 1, so 0:= Ii z+2 . qed Problem 4.39 Foran ideal gas, N = Avagadro's number/22.4liters = (6.02 x 1023)/(22.4x 10-3) = 2.7 X 1025. No:/€o = (2.7x 1O25)(47r€0x 1O-3O),8/€0= 3.4 X 10-4,8, where,8 is the number listed in Table 4.1. H: (3= 0.667, No:/€o = (3.4 x 10-4)(0.67) = 2.3 x 10-4, Xe = 2.5 X 10-4 } He: (3= 0.205, No:/€o = (3.4 x 10-4)(0.21) = 7.1 x 10-5, Xe = 6.5 X 10-5 .. Ne: (3= 0.396, No:/€o= (3.4x 10-4)(0.40)= 1.4x 10-4, Xe= 1.3X10-4 agreementISqUItegood. Ar: (3= 1.64, N 0:/ €o = (3.4 X 10-4)(1.64) = 5.6 x 10-4, Xe = 5.2 X 10-4 88 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER Problem 4.40 (a) (u) - J~:E ue-u/kT du - (kT)2e-u/kT [-(u/kT) - l]I~~E JPE e-u/kT du - -kTe-U/kT lpE-pE -pE { [e-pE/kT - ePE/kT] + [(pE/kT)e-pE/kT + (pE/kT)ePE/kT] } = kT e-pE/kT - epE/kT [ ePE/kT + e-PE/kT ] ( PE )= kT - pE epE/kT - e-pE/kT = kT - pE coth kT . A - - -(u) I { ( PE ) kT }P = N(p); p = (pcosO)E = (p. E)(E/E) = -(u)(E/E); P = Np pE = Np coth kT - pE . Lety ==P/Np, x ==pE/kT. Theny = cothx-1/x. Asx --+0, y = (~+ f - ~;+... )-~ = f-~; +... ~ 0, so the graph starts at the origin, with an initial slope of 1/3. As x --+00, y --+coth(oo) = 1, so the graph goes asymptotically to y = 1 (see Figure). .E... np' 11 """""""""""""'" . pe/kT (b) For small x, y :::::::kx, so;; :::::::-f!-r, or P :::::::~E = €oXeE =>P is proportional to E, and Xe = Np2 .p ~~ For water at 20° = 293 K p = 6.1 X 10-30 em' N = molecules= moleculesX molesX !\rams., 'volume mole gram volu e N - (6 0 10 23 ) ( 1 ) (10 6 ) - 0 33 1029. - (O.33Xl029)(6.1Xl0-30)2 - j"1;)l12 T bl 4 2 . - . X X 18 x-. X , Xe - (3)(8.85xl0-12)(1.38XlO-23)(293)- ~ a e . givesan experimental value of 79, so it's pretty far off. For water vapor at 100° = 373 K, treated as an ideal gas, v~~r::e= (22.4 X 10-3) X (~~~) = 2.85 X 10-2 m3. N = 6.0 X 1023 2.85 X 10-2 = 2.11 X 1025. , (2.11 x 1025)(6.1x 10-30)2 - Xe = (3)(8.85 x 10-12)(1.38x 10-23)(373)= 15.7x 10 3.1 Table 4.2 gives 5.9 x 10-3, so this time the agreement is quite good.
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