{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Introduction to Electrodynamics - ch04

# Introduction to Electrodynamics - ch04 - Chapter 4...

This preview shows pages 1–4. Sign up to view the full content.

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) = 7.34 X10-41. 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 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 0 9) x (2cos 0 f + sin 0 9)] 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)
75 . p2 sin 20 But sin 0 cos 0 = (1/2) sm 20, so I N = 4m:o(16z3) (out of the page). For0 < 0 < '!r /2, N tends to rotate p counterclockwise;for '!r/2 < 0 < '!r,N rotates p clockwise. Thus the stableorientation is perpendicular to the surface-either t or ..t.. Problem 4,7 Say the field is uniform and points in the y direction. First slide p in from infinity along the x axis-this takes no work, since F is J.. dl.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 16

Introduction to Electrodynamics - ch04 - Chapter 4...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online