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Unformatted text preview: Problem 3.1
The argument is exactly the same as in Sect 3.1.4, except that since 2 < R, V22 + R2— 22R =(R — z), instead of (z  R). Hence VWe = 7360* 22 R [(z + R) (R — z)]—  If there is more than one charge
411‘. cerI'IC inside the sphere, the average potential due to interior charges 1s—— 41re R and the average due to exterior
o Charges is Vcentery SO Vave‘” ‘ center + jgﬁcﬁ '/ Problem 3.3
Laplace’s equation in spherical coordinates, for V dependent only on r, reads: ‘72 — 7‘2 __ Q __ _ _ => V— __ Ci
=r—i dr (r2 d!" M) 0 :} dr *0 (conStant) d1" 1'2 Example: potential of a uniformly charged sphere. 1 d dV dV dV c
In cylindrical coordinates: V2V = — s—— z: 0 => 8—— : c=> —— = — => V = clns + k.
3 ds ds ds d5 3
Example: potential of a long wire.
Problem 3.6
Place image charges +2q at z = —d and ~—q at z : ~3d. Total force on +q is q ~24 2q —q  (12 1 1 1  29g .
F = . + . = —— — — — = ——
471'60 [0‘02 (4dl2 + (5d)2] z 47rfod2 ( 2 + 8 36) 2 41m; (72(12) z Problem 3.7 (a) Mom Fig. 3.13: a = \/T2 + a2 — 2m c050; 4’ = s/r2 + b2 — 2Tb c050. Therefore: (1’ R q _R‘
= _ ______i__.___. = _ 7
(ﬁ) 7'2+§; ~27'R72c050 (9]?)2lR2—21‘ac080
Therefore:
V(r (9):1 — —————————  ————l————— . 4—7r60 — 2m c080 R2 + (Ta/R)2 — 2rac039 Clearly, when r—  R, V —> 0. (b) a — "60 3—: (Eq. 2.49). In this case, %— — T at the point r— — R. Therefore, (7(6) _ q *1 .2 2_ —3/2 _ _ ‘
60 <47l’60){ 2(1 +a 2rac056) (2r Qacosﬂ) 1 2 2 3/2 (12
+ i (R + (ra/R) — 2racos 6) EQT — 2a c050 r=R ll 02
—4—q—7r{— (R2+a.2 —2Racos€) 3”(R R—acosﬂ)+(R2+a2—2Racos€) 3/2 (%—acos())} ll 2
4—:(122 +0.2 — 2Rac030)‘3/2 [R— ac036 — 2R +ac050] q 2 _ 2 2 2 _ ‘ —3/2
41rR(R a )(R +a 2Rac058) . ll ll /oda— _ 41—ng — a )/(12‘~’ + a2 — 2Racos€)‘3/2R2sinﬁd€ d¢ (Iinduced N H L022 _ (1.2)27I'R2 [__R1;(R2 + a2 _ 2Racos9)_l/2] 47rR 0
z 1(02422 [_____L____ ____1_____]
2a VIP + a2 + 2R0 R2 + a2 — 2R0 But a > R (else q would be inside), so VR’ + 02 — 2Ra 2: a — R. q l 1 q __ q
—(a2 122) [mm — (a_ ml: 20 [(a—m— (a+R)]  55—212) H H
l
l
H
Q (c) The force on q, due‘to the sphere, is the same as the force of the image charge q’, to wit: 1 qq’ 1 (0,1292) 1 __ 1 (12120 F” mom—b)? " 252; “ (a—Rz/aV ‘ 4m, (dz—122)? To bring q in from infinity/aI to a, then, we do work ————d2i——— qu 1 1 “ _ 1 (123
zzweooo (02 ~aR2)2 41r€o 2(62R2) 0° _ 41rco 2((22 R2).
Problem 3.8
Place a second image charge, q” , at the center of the sphere;
this will not alter the fact that the Sphere is an equipotential, 01,
but merel increase that otential from zero to V  ~—1——1 o—+——————
y P  4"“ R . q” q: q W
q" :2 41reoV0R at center of sphere. a For a neutral sphere, q’ + q” ' F . ; q;—:+ q, _ qql ‘i+ 1
_ 47reoqa. +—(a b)2 _ 41m, +—(a b)2 _q_q'_ b<2a — b) _ q(—Rq/a)'(R2/a><2a — Eva)
47reo a2 (a  b)2 ‘ 47reo a2 (a — Rz/a)2 = _ .21(£)3<2a2R2>
a 47m) ((12 — R2)? ll (Drop the minus Sign, because the problem asks for the force of attraction.) Isroblem 3.9 (a) Image problem: /\ above, —/\ below. Potential was found in Prob. 2.47: z v )=”1<s/)=*1n(si/s2)
y (y,2 41:0 n 3+ 41reo +
A 2 ln{ 3;2 + ____(______z + )2}
a: 41m) {+312 +(z — )2}
6V 8V 6V
_ .__ .1: —— t t "‘
(b) a: 60 871' Here—— an Bz’ evalua ed a z—
A 1 1
=  ——————————2 d  ——————2 z  d }
”(3’) 6°41reo {y2+(z+d): (2+ ) MHz—d)2 ( ) z=0
_ 2A d __ —d } _ _ Ad
 41r y2+d2 3/2lct2 — 1r(y2+d2)'
Check: Total charge induced on a strip of width I parallel to the y axis:
00 .
(Ad 1 _ [Ad 1 ‘1 y °° __l_é_¢_1~£____7:
(End = ”T wad'1"? lam“ (ﬁll; 1r 2 ( 2]
= —,\l. Therefore kind 2 —A, as it should be.
Problem 3.10
The image conﬁguration is as shown. y?
_ ___ . . _ . ..___.... —£1. » Eu mg (I
q 1 1 i
I” I — __;;______ __‘_.:_ *' i
( y) 4 0{ (I—a}2+(Jb) +z~ \/($+a)~ ——"'—f"’“
___ l _ l (1' ......... l ......... ;_._q
I For this to work, 0 must be and integer divisor of T869 Thus 180°, 90°, 60°, 45°. etc. are OK, but no others. It works for 45°, say, with the charges as shown. 45° line
(Note the strategy: to make the :1: axis an equipotential (V : 0),
you place the image charge (I) in the reflection point. To make the
45° line an equipotential, you place charge (2) at the image point. .7:
But that screws up the :2: axis, so you must now insert image (3) to
balance (2).l\"1(_)reover, to make the 45° line V = 0 you also need (4), 2. .
to balance (1). But now, to restore the r axis to V = 0 you need (5) (y, (E)
., . .,
to balance (4), and so on. why it works for 9 _ 450
. , . . . :0 l' ’ K/No good
The reason this (loesn t work for arbitrary angles is that you are even 130 Wile—10)
tually forced to place an image charge within the original region of (3”
interest, and that’s not. allowedrwall images must go outside the re— “j x
O gion, or you’re no longer dealing with the same problem at all.) 92“) "5(2)
why it doesn’t work for U —‘ 135° ...
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 Fall '09
 Lamb
 mechanics

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