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hw8solution - Problem 3.16 II I P3(1 AIH...

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Unformatted text preview: Problem 3.16 II I... P3 (1') AIH Oolwoolu—‘oo (5:1:3 - a: + 5x3 - 52:) = We need to show that P3 (cos 0) satisﬁes 1 d , dP _ . _ ma (sanQE) — —l(l + UP, Wlth l - 3, where P3(cos 0) = §cos€ (5 cos2 0 -— 3) . % %[— sin9 (5cos‘0 — 3) +cost9(10c030(— sin0)] = —%sin9 (500329 _ 3 + News” 9) = —gsin9(5c0820—1). % (31119513) = _gd£9 [sin20(5c0320 - 1)]: -% [231n0c056(5c0320— 1) + sin20(-10cosﬂsin6)] = —33in0c039 [5cos20 ~1- 5sin20] . —-l-—%(sin0(-i£) = —3c039[5c032-l—5(1—c0820)]=-3C059(10°0329‘6) l = .3.4.2 c039 (.Scos2 0 — 3) = —l(l + 1)P3. qed l l [PKWEMINz = fa); (5493 '33) d3 = g (2:5 ~23)|‘_l =%(1— 1 +1 - 1) :0. J -1 -l Problem 3.17 00 (3.) Inside: V(r,0) = Z Aw’P;(cos 9) (Eq. 3.66) where (=0 (21+1) " 2R’ Vo(0)Pg(cos 9) sin9d9 (Eq. 3.69). A.= 0 In this case %(0) = V0 coma outside the integral, so 1? A; = WQ/Hkosmsinow. o But Po(cos 0) = 1, so the integral can be written / Po(coso)n(coso)sinsda ={ ‘2" g :fg } (Eq. 3.68). 0 Therefore 0, if I 95 0 A‘={ v0, ifl=0 } Plugging this into the general form: V(r,9) = A0 roPo(cos 9) = The potential is constant throughout the sphere. 00 B! Outside: V(r,9) = WHROSO) (Eq. 3.72), where [=0 3, = (2’; ”12'“ / %(0)P¢(cosl9)sin0d9 (Eq. 3.73). 0 1f __ (21+1) 1+1 . __ 0, iflaéO _ ———2 R Vo/P)(cos9)sm0d0— 12%, “(:0 . o R . _ 1 Therefore V(r,0) = V0: (1.8. equals V0 at r = R, then falls off like ;). (b) 00 Z A,r‘1>.(coso), for r g R (Eq. 3.78) V0.9) = ’59 B . Z ;,—+‘—1P,(coso), for r 2 R (Eq. 3.79) [:0 where B, = R2‘+‘A, (Eq. 3.81) and 1 17 A1 = 2601214 /ao(0)H(c039)si110d0 (Eq. 3.84) 0 ﬂ 1 . _ 0, if I 79 0 260R‘_100/Pl(c050)sm9d9 —{ Rao/Eo, if I = 0 } 0 Therefore Note: in terms of the total charge Q = 41rR200, V(r. 9) = Problem 3.18 170(0) = kcos(39) = I: [4 cos3 0 — 30030] = k[aP3(c030) + ﬁP1(cos0)]. (I know that any 3rd order polynomial can be expressed as a linear combination of the ﬁrst four Legendre polynomials; in this case, since the polynomial is odd, I only need P1 and P3.) 4c0530 — 3cos€ = a [% (5c0530-300s0)] +ﬂcoso = 5§c0330+ (ﬂ _ go) c030, so So 8 3 3 s 12 12 3 “’2‘“? ””7“”—§'§=”‘3‘=ﬁ=?'3=‘3- Therefore I: Vo(0) = 3 [3P3(cos 0) - 3P, (cos 9)]. Now W ZAzr'mcosl’), forrSR (Eq. 3.66) V(r,0)= ‘5? B , ‘er'lmcosm. forrzR (Eq. 3.71) =0 where 21 1 ' A: — ( 2;, ) / vo(o)P.(coso)sinodo (Eq. 3.69) o z 1 k " " = (22;! )3 {s/Ps(cosﬁ)1’1(c050)sin9d0-3/P1(cosa)P1(cosﬂ)sin9d0} 0 o _ k(21+1) 2 2 _ 5i _ _ 5 23‘ {8(21+1)6‘3 3(2z+1)6“} ' 5Rl [86‘s 3‘5“] 8k 5123, 11:3 , = { —3/k/5R, lf I = 1 }(zerootherw13e). Therefore VOW") - ‘5—RTP1(0069) + ﬁr P303080) — 5 [8 (R) Pa(coso) —3 (R) P1(c060)], 01' He (1%): % [5cos30-3c080] —3 (i2) c089} => V(r,0) = §%cosq{4 (ﬁr [5c0820— 3] F3} (for r g R). Meanwhile, B, = AIR”+1 (Eq. 3.81—this follows from the continuity of V at R). Therefore 8,6124 52 If I z: 3 ' BI = { ‘3kR/2 / 5, if I = 1 } (zero Otherwise), So .. 2 4 32R: —P1 (cos 9) + 82R: V(r, 0): —-P3 (cos 0): [8 (R)4 P3(cosﬁ)- 3(§)2 P1(c050)] , V(r,0)= §(R)26089{4 (§)2[500s20—3]—-3} (for 1' Z R). Finally, using Eq. 3.83: UHF? 01‘ 0(9) = 502(2l+1)A1R"1P,(cos0)=eo[3A1P1+7A3R2P3] 1:0 = 60 [3( 53%) P1+ 7 (53;) 122133] = 595 9P1( (coso) + 56P3(cosl9)] _ 60k 56 60 k —— —5—-R[ 9cos€+ 2 (5cos3 0—- 3c050)]= -5——°Rcos0[— —9+28- Secs2 0- 28- 3] 60k = —5I—zcosl9[140cos2 0-— 93]. Problem 3.19 °° 21+1 I Use Eq. 3.83: 0(0) = so Z(2l+1)AgR"1P;(cosB). But Eq. 3.69 says: A; = W/%(6)P1(c050)sin9d0. Putting them together: 0(0):: -2€-°I§Z(2l + 1)2C'1P;(cos 0), with Cl: / V0 (0)P;(cos 9) sin0 d9. qed =0 Problem 3.20 Set V = 0 on the equatorial plane, far from the sphere. Then the potential is the same as Ex. 3.8 plus the potential of a uniformly charged spherical shell: R3 1 Q V(1‘,9) — ~Eo (1‘ - :5) 6059+ m7. Problem 3.22 00 ZA;r‘H(cos€), (r g R) (Eq. 3.78), V(r.9) = ‘3‘? B Z 1.1+! -—P;(cos€) (r 2 R) (Eq. 3.79), l: where Biz/1‘12”“ (Eq. 3. 81) and 1f 1 n A! = 260R“,[00(9)P1(c089)sm9d€ (Eq. 3.84) o Ir/2 1r 2 26011131400O/PNCOSgNUWda-/Pl(c059)sin9d0 (letz=cosg) 1r/2 2601214 {/H(x)dz—/0Pl(z)dz}. Now I’d—z) = (—1)‘P¢(z), since P;(z) is even, for even I, and odd, for odd 1. Therefore 1 0 0 P102)!” = P1(-\$)d(-\$) = (-1)‘ PM) dz, 1 I / 0 and hence 0, if I is even 1—( —1 P A1: 260121— 1[ ) ]/ t(a:)d 60R,_°lo/1P,( 3:)dz, iflisodd 80 A0 = A2 : A4 = A6 = 0, and all we need are A1, A3, and A5. 1 l /P1(z)da: = fzdzzg 2. 0 0 0 1 1l 1 \$4 .52 l 1 5 3 1 — — 3-— = — '—— — —— : — — — — :-—— /P3(x)d:c — 2/(5:c 32:)da: 2(04 32)0 2(4 2) 8 0 O l 11 1 b 4 2 1 /P5(a:)dz = —/(63:c5—70z3+152:)dz=—(63\$——7OI—+15£— 8 8 6 4 0 0 0 1 21 35 15 1 1 - §(?‘3+7>‘E‘36*35) — Therefore and Thus [P1 (cos 6’) [P1 (cos 6) — 260T2 Problem 3.26 Monopole term: Q =/pd1' = 1:12] [\$(R— 2r) sinHJ 'r2 sin9drd0d¢. But the r integral is n /(R—2r)dr= (Rr—r2)|:=R2—R2=0. SoQ=0. o Dipole term: / r cos0pd‘r = H? /(r cos 0) [1712-02 — 2r) sin 0] r2 sin 0dr d0 d4). But the 0 integral is /sin20cosﬂd0 = 0 So the dipole contribution is likewise zero. Quadrupole term: “/‘r2 (2 00820 — %) pdr = ékR/[rz (3c0320- 1) [1%(12— 2r)sin9] 1'2 sinﬂdrdo. . I sm3 0 3 (0-0)=0. l o 3 r integral: R 3 4 2 _ = T. -7". A r(R 2r)dr (3R 2) o 3 2 6 6 integral: I X X / (3c0320 — 1) sin29d0 = 2/sin20d0 - 3/sin‘ode ‘—sr-—’ ° 3(1-8in’0)—l=2—33in30 0 0 w 371' 9 1r - 2(a)"?‘(ﬂ ”(1—5;)"? 45 integral: The whole integral is: 1 R4 1r k1r2R5 MT) <-§><2«>- 48 - For point P on the z axis (1' —> z in Eq. 3.95) the approximate potential is 1 k1r2R5 5 _ V(z) — 41reo 4823 ' (Quadrupole) Problem 3.27 p = (3qa - go) 2 + (-2110 - 2q(—a))\$r = 2qai. Therefore and p«f' = 2qai-I“ = 2qac050, so 1 anc039 . N _... Problem 3.28 (a) By symmetry, p is clearly in the z direction: p = p2; p = f zpdr => f 20 da. ‘8' " 3 p = [(Rcos0)(kcos9)R3sinad0d¢ = 21rR3k/cos205in0d0 = 21rR3k (3°53 9) 0 0 41rR3k 41rR3k. 3 ; p: 3 z. Eli-«R3141 — (—1)] = (b) V“ l 41rR3kcosﬂ_ Ecosﬂ —41reo 3 r2 350 7.2- (Dipole) Problem 3.31 . . . p a p A m *- : 9 = E = = = _ I = = _ (a) This pomt Is at. 1- a,, 2, gt: 0, SO E 41reoa3 9 41r£oa3 ( 2); F (IE 41600.3 2. (b) Here 1' = 0, I9 = 0,80 E 2 47mm} (21") — .. _ = 913 (C) V _ q [V(OI 0! a} V(a! 0) 0)] 47(60112 Proble— "3 32—_ 1 1 0 Q = —q, SO Vmono : __Q; P = 90 2a 30 Vdip = __qacos . Therefore 130,9) 2 ——— —if+ i (2cosai+sinoé)]. 41m) r2 r3 Problem 3.33 - .. . . - ‘ .. p = (p-i')1"+ (p-9)0 =pcost—psin60 (Fig. 3.36). So 3(p-f)f—p = 3pcosGr-pcosﬁr+psnn90 = 2Dcos0f' + psinﬁa. So Eq. 3.104 E Eq. 3.103. f ...
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hw8solution - Problem 3.16 II I P3(1 AIH...

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