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Unformatted text preview: 3.4. In cylindrical coordinates, let D = (pa, + 28; [41r(pa +z’)“°]. Determine the total ﬂux leamng:
a) the inﬁnitelylong cylindrical surface p = 7: We use oo 91!
_ _ M 2 L
“’“ID'JS‘LO/o wanna/9'“'”°d¢d"”°[(pa+za)w where p0 = 7 (immaterial in this case).
b) the ﬁnite cylinder. p = 7, z g 10:
The total ﬂux through the cylindrical surface and the two end caps are, in this order: to 21v
_ ML
tbs—[20]) 4xW+za)a/9md¢dz
2" 0° zoazae 2' '° ma:'&'.
+ —,— dd +/ / —,— d d
[o [o 4"(P9+zo)3/2PP¢ o o «(puma/9”” where p0 = 7 and 20 = 10. Simplifying, this become db [:3 0 —,—d3 + 20 M —5—de
b _
0 (Po + 29)“ o (P’ + 20)”
_ z :0 _ 20 P0 _ m + 1 _ 20 _ l
—‘/;_73—7+z 0 7m 42—0 “ﬁg—3+? —¢m  where again, the actual values of po and so (7 and 10) did not matter. 35. Let D = 429a, + 2(::9 + 2%, + 45m, 0/113 and evaluate surface integrals to ﬁnd the total
chargeenclosedinthe rectangularparallelepiped0<1<2,0<y< 3.0< z< 5m: Ofthe6
surfaces to consider. only 2 will contribute to the net outward ﬂux. Why? First consider the
planes at y = 0 and 3. The y component of D will penetrate those stufaces. but will be inward
at y = 0 and outward at y = 3. while having the same magnitude in both cases. These ﬂuxes willthuscancel. Atthex=0plane,D, =0andatthez=0plane,D,=0,sotherewillbe
no ﬂux contributions from these surfaces. This leaves the 2 remaining surfacm at :r = 2 and
z = 5. The net outward ﬂux becomes: 3 3 2
6:];[0 leagdydzF/o lo Dlzdazdtdy
=5/34(2)ydy + 2/34(5)ydy=360C
O 0 3.7. Volume charge density is located in free space as p, = 2e"m°' nC/rn3 for 0 < r < 1 mm, and
p, = 0 elsewhere.
a) Find the total charge enclosed by the spherical surface r = 1 nun: To ﬁnd the charge we ' tegr t :
m 88 an It 001
0:] f / 2e'1m’rnsin0drd9d¢
0 O 0 Integration over the angles gives a factor of 41f. The radial integration we evaluate wing
tables; we obtain 4261000.
Q = 3" [W willthuscanceL Atthez=0plane,D, =0andatthez=0plane,Dz=0,sotherewillbe
no ﬂux contributions from these surfaces. This leaves the 2 remaining surfaces at z = 2 and
z = 5. The net outward ﬂux becomes: 3 3 9
¢=/:/; Dlpaﬁdde'l'fo [o Dlzdazdzdy
9 ‘I .ml 2 e—lmﬂ'
o + 1000 (1000)? .ml 9
(—1000r—1)o ] =4.0x 10 nC 3.17. A cube is deﬁned by l < 1',y.z < 1.2. If D = 2223's, + Myaag C/m’: a) apply Gaus‘ law to ﬁnd the total flux leaving the closed surface of the cube. We call the
surfaces at 1' = 1.2 and 1' = l the front and bad: surfaces respectively. those at y = 1.2
andy: 1 the right andleftsurfaces, and thoseatz =1.2andz = l the topandbottom
surfaces. To evaluate the total charge. we integrate D  n over all six surfaces and sum
the results. We note that there is no 2 component of D, so there will be no outward ﬂux contributions from the top and bottom surfaces. The ﬂuxaa through the remaining four are
L9 L9 L2 L9
o=Q=fDmda=j / 2(1.2)’ydydz+/ / —2(1)’ydydz
l l l l ‘ﬂ—JVﬂ—J
hum but L2 L9 L9 12
+/ / _3z3(1)9dzdz +/ / 329(1.2)’dzdz = 0.10282
1 l 1 1
\________‘,________J \________"________J
m right b) evaluate V  D at the center of the cube: This is
V D = [42y +622y]0_1.1‘1] = 40.1)? + 60.1)3 = 12.83 c) Estimate the total charge enclosed within the cube by using Eq. (8): This '3
Q 5 V  DIM“,r x Av = 12.83 x (0.2)3 = 0.10215 Close! 3.25. Within the spherical shell, 3 < r < 4 In, the electric ﬂux density is given as
D = 5(r — 3)3a. C/m2 a) What is the volume charge density at r = 4? In this case we have A, =v . D = ggﬂx) = 20 _ 3)9(5r _ 6) 0/1113 which we evaluate at r = 4 to ﬁnd “(7 = 4) = 7.50 Clans. b) What is the electric ﬂux density at r = 4? Substitute r = 4 into the given expression to
ﬁnd 0(4) = sa,C/m9 c) How much electric ﬂux leaves the sphere r = 4? Using the result of part b. this will be
«t = 4«(4)9(5) = 320w C d) How much charge is contained within the sphere. r = 4? From Gaus’ law. this will be
the same as the outward ﬂux, or again, Q = 3201f C. ...
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This note was uploaded on 11/28/2009 for the course EE EE 1 taught by Professor Ozcan during the Spring '09 term at UCLA.
 Spring '09
 Ozcan

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