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20091ee1_1_hw3-sol

# 20091ee1_1_hw3-sol - 3.4 In cylindrical coordinates let D...

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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 inﬁnitely-long 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° zoaz-ae 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 4-2—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 lea-gdydz-F/o lo Dlzd-azdtdy =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 426-1000.- 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 ¢=/:/; Dlpa-ﬁdde'l'fo [o Dlzd-azdzdy -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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