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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
48. Two point charges. (a) We 'can use the general equation [4.73 for the electric ﬁeld from multiple Fig. 4.3. Figure for Problem 48. (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; . 412. 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 1012 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. 413. Two straightline 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) Vm“ 4:
It [s 4:. 416. 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 416. which can be integrated to ﬁnd the total ﬁeld at the origin due to the halfring, 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 halfring 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). 418. 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 MATIIEMATICA). 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 419. 420. 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 3432—4337 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“ ebr 2K 1" 3”
<1: = ————— — + —— 5? m M __ .._
(T) E07" (53) (5052) e + 605 f r dr+ cabs / 12 dr where the last two terms are the socalled 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. 423. 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: .xmmeq 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 423. E. and (1) versus 1‘. 4M. 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 413, 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 424. 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.
 Summer '08
 STAFF

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