Introduction to Electrodynamics - ch04 - Chapter 4...

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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 = (7.34 x 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: J E.da = E~ Qenc, or E = 4;<0 ~Qenc. l r 47rq l 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~ + 2 a2 . [Note: Qenc(r --+ 00) = q.] So the fieldof 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
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74 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER 1 q ( 4 d3 ) 1 4 1. I 3 I E = -- -- = --(qd) = -po a = 311"!:oa . 471"€0 dl- 3 a3 471"€0 3a3 371"€oa3 [Not so different from the uniform sphere model of Ex. 4.1 (see Eq. 4.2). Note that this result predicts 4;EO a = !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) = -!o Qenc = Elo J; Ar471"r2 dr, or E = ~ 471" A r4 = Ar2 . This "internal" field balances the external field E when nucleus is "off-center" an amount 471"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
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This note was uploaded on 02/17/2010 for the course PHYSICS mae142 taught by Professor Kim during the Spring '10 term at American College of Computer & Information Sciences.

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Introduction to Electrodynamics - ch04 - Chapter 4...

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