HW 4 Sol - fiflifif Wadi/4 éc fi/K/ flea/car charges,...

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Unformatted text preview: fiflifif Wadi/4 éc fi/K/ flea/car charges, with the position vectors of the two charges Q1 = +62 and Q2 = "BQ being respectively given by r’; = 2? and 1"; = —lj>. The point we are to evaluate E is at r = 0, so that r — r’1 = ——2§ and 1' ~ ["2 = y. We have 2 E(0,0,0)=_1__ZQ_’°(£:_‘:‘Q= = Q 47r€o Il' ‘- r’kP 8 1 4 41r€9 i E 3 4-8. Two point charges. (a) We 'can use the general equation [4.73 for the electric field from multiple Fig. 4.3. Figure for Problem 4-8. (b) Let the point of zero field be at (0, y, 0). We know from the geometry that y > 2 or y < —-1. We have _ l Qty—2)? (“39)(y+1)9 _ E(°’y’°)‘2?r23[(y—2)3 l" (y+1)3 ]“° 1 __3,_____ _+ 2_ _ 2 _, y2—14y+11=0 m» y=0.900r6.1 Since the first solution does not make physical sense, the only nontrivial answer is y = 6.1. This makes physical sense since there must be such a point at which the repulsive force due to Q; is balanced by the attractive force due to Q2. Note that there does not appear to be a solution for a point located at y < —1 (Le, to the left of 62;); this makes physical sense since the attractive force due to Q; is much too strong to be canceled out by the repulsive force due to Q; . 4-12. Six point charges on a hexagon. (a) The electrostatic potential at the center of the hexagon is given by __ Q (I) — 6 (47mm where Q = 25 nC and a = 6 cm is the length of each side which is also the distance d from each charge to the center of the hexagon since sin 30° = a/(Zd) ——> d = a. Substituting the numerical values, we find —9 : 6x25x10 222.51“, 1 4w x 8.85 x 10-12 x 0.06 (b) The energy required to move a point charge of "25 11C from infinity to the center of the hexagon can be calculated as W = qe 2 (—25 x 10—9)(22.s x 103) z -5.62 x 10‘4 J Note that the minus sign indicates that the energy required to bring the —25 11C charge to the center from infinity is provided by the field of the six positive charges. 4-13. Two straight-line charges. (a) With p; = 100 nC/(I m) = 10‘7 C—m“ the potential at point P 3 g i i i i E E E t is given by (I) 1 /1 pl (i9 + /2 ,0; (135 P = .—-——— _._._—._ 4 t/ 2 2 -— 0.5 7‘50 o (0.5) + y 1 (a: ) val-[iced line Horizontal line 2 . 1 2 = pl { [ln(y + V 0.25 + '93)] + [111(51: — 0.5)] } 41T€0 0 1 1 + x/ 1.25 1.5 =9 ><109 ><10‘7 >< ln——-——+1n—— 22.29 x103V [ V0.25 0.5] (b) —_._.._-..—-—— E __ 1 ' (0.5:: ~ gym dy + 2 (0.5 — ans dm] 1’ '" 4m 0 [(0.5)2 + y213/2 1 (a; — 0.5)3 m l W60 / 0.5dy _y 1 ydy _R/2 :19: ] 0 (y2 + 0.25)3/2 0 (y2 + 0.25)3/2 1 (a: — 0.5-)2 1 4 2 ) 4] 2—— — 2 m — — s— 7“50 [ x/5 ?( J5 3 ~-7 9 M) (3—4)] _10 ><9><10x[(\/g 3 R+ V3. 9 g (410;: w— 995 y) V-m“ 4: It [s 4:. 4-16. Circular ring of charge. (a) The potential at the origin is given by ‘92 _QL_ 41reoR + 47reo(2R) _/”szd¢'+ Q2 1+ 1 _ 0 47ng 41769 R 2R _ 1 pter L02 _ 1 3:22 _41reo( R +2R h47r£oR Q14. 2 where we have noted that p; = Q1 /(1rR). Thus, (I) = 0 requires Q2 :- —2Q1/3. (b) Noting that, due to symmetry, the half—ring electric field at the origin has no 3,: or 2 components, so that we can simply work with the a: component of the field. The 2: component of the elemental electric field at the origin due to a charge element pgdl’ along the half ring is given by pzdl’ . pzdl' . l’ dE = = .- [ mielemem AtmoRz Sing 41m) R2 sm R (I, = (phalf ring + .Fig. 4.7. Figure for Problem 4-16. which can be integrated to find the total field at the origin due to the half-ring, i.e., 1r 7' ’ pl . l ZQI/W E . = = _ I = [ xlhair flag /0 4MOR2 f0 Sm (3) (fl 41reoR2 so that the total fieid at the origin due to both the half-ring and the two charges is 47reoR2 — 47mng _ 47reo(2R)2 47reoR2 _ m H E(0,0)=2Q‘/”sk Q2 2 ——————QZ 2:2 1 [23‘ 532] Setting E = O we find Q2 = 8621 /(51rr). 4-18. Charge on a hemisphere. (a) Considering the geometry shown, and using [4.28] with [1" I = a, r = zfl and R = [r — r’l = Va? + 2:2 — Zaz cos 6’, the electrostatic potential at point P is given by (D m [7* f2" pads’ __ om) 1" a2 sin s'da' P 9’=1r/2 ¢r=o 47"EOR (2W02X47T60) «[2 Va? + 22 — Zaz cos 6’ where we have noted that p3 = Q/(47ra2) and ds’ = a2 sin H’dfl’dqfi’ . The integral can be evaluated by substitution of variables (e.g., let u = a2 + a2 w Zaz cos 6’, in which case du = u sin 6’038’) or by looking up in a Table of Integrals or by other means (e.g., using a mathematical analysis package such as MATI-IEMATICA). We find Q [n+2 Val—+22] (PP 41reo az az (b) By symmetry, we expect the electric field at point P to be in the z direction, i.e., Ep = 25(2). We can find E(z) by taking the negative gradient of (PP in the z direction, i.e., E(z)__6€D(z)__ l_a+z 1 + (9+2:2 _ 62 H az we2 m/az + 32 Liz2 (c) We follow the same procedure, except that we now consider a volume charge density p = 3Q/(21ra3) and also integrate over 1". In other words, we have -Mmewwmwwwmmmmmwmew- ii 362(271') “ /" («HP sin 6’d6’dr’ (DP Ti‘ = (ZWGBX47F60) rr=o /2 Va2 + 31 — Zaz cos 6’ where the integration Met (15' is already done. Looking up the integral from tables or using an analysis package we find 362 z2 + a2(2a, + 32) _ (a2 + 2:2)3/2 ‘13P = 47rega3 3 62 3'2 The electric field Ep = fiE(z) and E(z) can once again be found as the negative gradient of (DP in the z direction. We find 2 23/2 (1 +2) + a2+z2———————( 32: 47reoa3 3 22 622 322 Fig. 4.9. Figure for Problem 4-19. 4-20. Spherical charge distribution. (a) The total charge Q in the spherical region 0 < r < a. is given by 211‘ ir a I 2 Q = p(r’)d'v’ = K/ f / 6‘4"" 1" sin B’dr’dfi’dgb’ V' 0 0 0 2 =K(21r)(2) — 6"“ (9;— + g; + where for the r’ integral, we used 2 fuze‘mdu = ea” [3— — E + which can easily be shown using integration by parts twice. 3.] W .. W WW WW we; W we Wimmmummmmmmmmnwwmmmmv MW mmmm 5W3.» J rm“ m n m x. mm m. WA” :>\ 400‘ W M A n WWWKMMI (b) Since there is spherical symmetry, we consider a spherical Gaussian surface S with radius r and apply Gauss’s law for the cases when r < a and r > a with the result of part (a) as 2 / 60E ' d5 = EOEr(47TT2) = Qm = (41rK) [wr— —— E: - s resulting in K 2 mbr 7'2 2T 2 agile—e (Warren 7‘5“ K 2 _ba 0,2 20. 2 '—"" 33—8 34-32—4-337 r>a (c) The electric potential can simple be found by integrating the electric field, namely 2 CD0") = —— / Er(r)d'r 00 For '1" > a. we simply find For T g a we have K 2 K _ 2K 1“ e-br 2K 1" 3-” <1: = ————— — + —— 5? m M __ .._ (T) E07" (53) (5052) e + 605 f r dr+ cabs / 1-2 dr where the last two terms are the so-called Exponential Integrals for which there is no closed form solution but which are well tabulatedi. (d) The result can be shown by simple substitution into the cylindrical coordinate version of Poisson‘s equation, namely 2 2 _ ‘W" 0 i No variation in 45 N0 Valiation in z 1 a ( 8(1)) _ FKe—br $6? T5; 50 Note that differentiation of the Exponential Integral terms simply yield the integrands. 4-23. Spherical charge distribution. (a) The total charge Q can be calculated as 22¢ 11' a 2 Q = p(r’)do’ = f / f p(r’)'r’ sin H’dr’dfi’dgb’ v' o o o a. :2 :p0(2)(27T) f0 (1— r'zdr’ (b) Due to spherical symmetry, th an'd 'depend only on r. We cons' ongm and apply Gauss’s law: .xmme-q ewmwmwrwwmw I 81rp0r3 % 60E ds = €0Er(47rr2) = Que _ 15 T a S 37rpoa3 15 7' > a g l a i i 5% g i E i i a E E (c) The electric potential can be found from the electric field as r (110') = -~ / E - oil 00 For r > a, this integral results in E r (110') a — [m 29°”? dr = _2'°°“3 fr Eli = Mas l r = zpoaa 0° 1560r2 1560 00 r2 1550 r 00 3 r<a 2 2 2 _ 2 (MT) =2420a __ / pgrdr = 2pm 2pc r a ISGOG. a 15w 3 For T < a, this integral can be evaluated as l l I l l l I a Fig. 4.10. Figure for Problem 4-23. E. and (1) versus 1‘. 4-M. Spherical charge with a cavity. We can consider the cavity region, within which the charge density is zero, to result from two equal but opposite charge densities p1 = K and p2 = —K. Accordingly, we can represent the problem in hand as a superposition of a uniform spherical charge distribution of radius b with charge density ,0; = K without a cavity and another of radius a. with a charge density p2 = —K respectively. From Example 4-13, the electric field at point P(r, 6, qb) inside the spherical region of radius a due to the uniform charge distribution ,0] = K in the spherical region of radius (2 can be written as E]: Q1 4 ,_ _ 3 4fleob3 r: where Q1 _ Barb K Similarly, the electric field at point P(r, 6', 925) due to the uniform charge distribution p = -—K in the spherical region of radius a follows as _ Q2 _ I 3 _ h __ __. E2 31‘2 W ere Q2 1m K Fig. 4.11. Figure for Problem 4-24. Spherical charge distribution with a cavity. Therefore, the total electric field is K E =E1+E2 = “(r1 — 1'2) 360 Without loss of generality we can choose the center of the hole to lie along the z: axis of a coordinate system having the center of the larger sphere as its origin. With that choice, and also choosing the point P to be along the 3: axis, we have r1 — r2 2: it'd, so that E = [It'd/(360ml. ...
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This note was uploaded on 01/17/2011 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.

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HW 4 Sol - fiflifif Wadi/4 éc fi/K/ flea/car charges,...

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