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# hw6solution - Problem 3.1 The argument is exactly the same...

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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]?)2-l-R2—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(62-R2) 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, 0-1, 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<2a2-R2> 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 3/2-l-ct2 — 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+(J-b) +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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