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Unformatted text preview: Problem 4.2
First ﬁnd the ﬁeld, at radius r, using Gauss’ law: I Eoda = «‘10 ch, or E = 337; F1303“. 7' 7' 41”! 8—2770 2 4Q a _2r/¢ (12
0°“ _ Lpdr=msr o :di‘=—3 ——e r2+ar+2 0 2
~2—q 6‘2”“ 7‘2 +ar+— —— =q l—e'zr/a 1+2£+2L .
a2 a a? {Note Qencb‘ a co) — q] So the ﬁeld of the electron cloud IS Ee= —— [I — 62”" (1 + 25 + 25)]. The 41m) proton will be shifted from r = 0 to the point d where Ea: E (the external ﬁeld): E: Li [1 —e"“/° (1 +23 +23%”. 4176052
Expanding in powers of (d/a):
2 2 3
a 2 a 3! a a a 3 a
2 3 2
.1—eW“ 1+29+2d2 = 1— 1'23” 5 3 9 + 1+2§+2d—2)~
a 0.2 a 3 a a a
d2 (13 4d3
= y 2¢ 2§+2g+4g+4iﬂ 26¢" 4$+5§+H 1’
= g (g)3 + higher order terms. 1 q 4 d3 l l
— — d .
47r€o d2 (3 (2—3) 47reo (303‘! )= 37reoa3 p a 37reoa [Nlot so different from the unifoarm sphere model of Ex. 4.1 (see Eq. 4.2). Note that this result predicts "(0a = 2a3 = — 3.(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 3° m3 Ironically the‘ ‘classical” formula (Eq. 4.2) is slightly closer to the empirical value]
Problem 4.4
r Field of q: ﬁgﬁrf‘. Induced dipole moment of atom: p = aE =
5 9 5:31;; 1'.
Field of this dipole at location of q (0 — 7r, in Eq. 3.103): E: —1—i ( 2aq ) (to the right).
’ 47reo r3 47rosor2 2
1
Force on q due to this ﬁeld: F: 20 _q__ (attractive).
47m) 1'3 ) rs 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 .L :11. 1E (If E is not uniform, slide p in along a trajectory .L the ﬁeld.) Now 9 rotate (counterclockwise) into ﬁnal position. The torque exerted by E is N = pr = pEsinBi. The torque we exert is N = pEsinﬁ p a: clockwise, and d9 is counterclockwise, so the net work done by us is
negative: U = If” pEsin§d§ = PE (— cos§):/2 =—pE (c080 — cos— 3“) — —pEcos€ — —p .E qed i) Problem 4.9
q mx+yy+zz (a) F = (p  V)E (Eq. 4.5); E 2 L12 r: 41m; 1‘ 41m) (x2 + y2 + z2)3/2
6 8 8 q z
F = ——  — ——““—
2 ( ’32 ““03, +p382) 41750 (x2 +y2+22)3/2 q { [ I _ gm 2x 3 29
4m '3‘ ($2 + y2 + 22)” 2(x2+ 92 + 22)“ +9” 5250:? + y2 + 22)“? 3 2 .
+ pz [—_I—.—w—i—]} : q [BE "‘ 133(p3$+pyy +pzz)] __ i [B _. 3r(p l‘)] . (.1:2 + y2 + 22)“2 4—1rco r3 41m, 13 r5
1
F :
47:0 1'7 ql—p 3(p r) r]
1 l ‘ ‘ , .
(b) E: R73 {3[p (—r)]( —r) — p} — 4m 670 ~—r—3 [3(p  r)r — p]. (This 18 from Eq. 3.104; the minus signs are because r points toward p, in this problem ) 1 F = E— ‘
q 4—7rco r_3 q[3(p r) r — 13] [Note that the forces are equal and opposite, as you would expect from Newton's third law.] Problem 4.18 (a) Apply f D  do = Q1,“ to the gaussian surface shown. DA = 0A => (Note: D = 0 inside the
metal plate.) This is true in both slabs; D points down. m”
u E(b) D: 5E =:> E: 0/5; in slab 1 E: 0/62 in slab 2. But 6 — 606,, so 61 = 260; 52 — gm (c) p = toer, so p = com/(cot) = (Xe/w; x, = 5, — 1 e p = (1 — Ema. (d) v = Ela + 132a = (ca/660)(3 + 4): = +P1 at bottom of slab (1) = 0/2, = +P2 at bottom of slab (2) = 0/3, (e) ‘0" = 0’ 0,, = —P1 at top of slab (l) = —0/2; = —P2 at top of slab (2): 0/3.
_ total surface charge above: 0 — (0/2) = 0/2,
(f) In slab 1' { total surface charge below: (0/2) — (0/3) + (0/3)  0 = —a/2, => E1: 22' J . total surface charge above: 0 — (0/2) + (0/2) — (0/3) = 20/3, 20
In Slab 2' { total surface charge below: (0/3) — 0 = —20/3. =WE2  3—50 I 2:: +a
——————————a/2 (D
+0/2
~0/3 C2) +0/3 :::—0 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  0/50 (in air) and E‘ — 0/6 (in dielectric). 80 V: %+ 11% 3"— (1+ 5:1). _ 2(oA
C 2:
__ __ “A 2 u; _a 7'
Ca’3‘d (1+1/¢,) 0—0 =1+€r. In conﬁguration (b), with potential difference V: E = V/d, so a = 50E = eoV/d (in air). P = <50er = (excl/7d (in dielectric), so ab = —eoer/d (at t0p surface of dielectric).
om : coV/d : a; + 0), : a; — cox,V/d, so a; = col/(1+ xe)/d = eoch/d (on top plate above dielectric). _Q_i ﬂ §__4_ v v _Aco 1+6, Cb_1+e,
:Cb”v‘v("2+U’2)“2v<‘°d+‘°d">‘7 2 "0—0— 2'  . 7 9A _ C _ iii: _ 2r _, 1+:' 2—4:, 1+2er+4e2— —4cr i—e. ’
[Which 15 greater. Co E: _— 2 H: — w: W: Sal—+33 > 0. So Cb > Ca.)
If the :1: axis points down: —I—:_m__
Il mln a?“ Enem—
l—  "un
—II er 2 w— —‘l_ Problem 4.20 fDoda = Q)“: => D41rr2 = p§7rr3 => D = §pr é E = (pr/3dr, for r < R; D47rr2 = p§7rR3 => D =
p123/3r2 => E = (pR3/3eor2)f, for r > R. “Fl o 2 2 p pR pR2 __p_R 1 __ d =__. ____ _
36/ r r 360 +36 2 3—60 (1+ 26,.) 00 Problem 4.21 Let Q be the charge on a length 8 of the inner conductor. waia ll D27rs€= Q:>D: Q; E: Q (a<s<b), E: Q (b<r<c). 27rs€ 27reos€ 21(583
“ " ds 0 Q d3 Q b 60 c
V =  E. : — —— —:= — —  .
fc d1 [a (27203) 8 +/; (2nd) 3 21m)! [In (a) + e ln(b)]
_C_ _ g_ 27(60
3 _ Vt? * ln(b/a)+(1/£,.)ln(c/b)' ...
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 Fall '09
 Lamb
 mechanics

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