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20115ee1_1_HW Chapter5

# 20115ee1_1_HW Chapter5 - 4.31 A potential ﬁeld in free...

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Unformatted text preview: 4.31. A potential ﬁeld in free space is expressed as V = 20/ (xyz) V. a) Find the total energy stored within the cube 1 < x, y, z < 2. We integrate the energy density over the cube volume, where 1115 = (1/2)60E - E, and where l l l xzyg ax + Iyzzay +— I—y22 E=—VV=20[ a]V/m The energy is now 2 2 2 l l l W = 2 E 0060]; j; 1; [111431222 + 111237422 + 3123334] dx 0'de The integral evaluates as follows: 2 7 1 7 1 1 1 #0060], [(a a+(a);+(2)a “’2 7 b) What value would be obtained by assuming a uniform energy density equal to the value at the center ofthe cube? At C(LS, 1.5, 1.5) the energy density is 1705 = 2006013) [ = 2.07 x 10—10 1/1113 (1.5)4(1.5)2(1.5)2] This, multiplied by a cube volume of 1, produces an energy value of 207 p1. 4.35. Four 0.8 nC point charges are located in ﬁ‘ee space at the corners of a square 4 cm on a side. a) Find the total potential energy stored: This will be given by where V,l in this case is the potential at the location of any one of the point charges that arises ﬁ'om the other three. This will be (for charge 1) q l l l _:| v=v v +v = _+_+ 1 21+ 31 41 47reo[.04 '04 .04ﬁ Taking the summation produces a factor of 4, since the situation is the same at all four points. Consequently, 1 (.8 x 10-")2 [ 1 ] _7 w =_4 V=—2 _ =7.79 10 J=0.779 J E 2( )41 1 27r€o(.04) +J§ x p, b) A ﬁfth 0.8 {.LC charge is installed at the center of the square. Again ﬁnd the total stored energy: This will be the energy found in part a plus the amount of work done in moving the ﬁfth charge into position from inﬁnity. The latter is just the potential at the square center arising from the original four charges, times the new charge value, or AW _ 4(.8x10'9)2 _ 813 J E _ 4ﬂ€o(.04ﬁ/2) _' u The total energy is now WE”, = WE(Parta) + AWE = .779 + .813 =1.59111 5.1. Given the current density J = —104[sin(23c)e'2yaJK + cos(2x)e'2yay] kA/mz: a) Find the total current crossing the plane y = l in the ay direction in the region 0 < x < l, 0 < z < 2: This is found through 2 l 2 l I=ffJ-n‘ da=f f J-ay‘ dxdz=f f —104cos(2x)e'2dxdz S 5 o o 7:1 o o 1 2 l = 7104(2)§sin(2x)‘ e' = 4.23MA 0 b) Find the total current leaving the region 0 < x,x < l, 2 < z < 3 by integrating J - (is over the surface of the cube: Note ﬁrst that current through the top and bottom surfaces will not exist, since J has no 7. component. Also note that there will be no current through the x = 0 plane, since JJr = 0 there. Current will pass through the three remaining surfaces, and will be found through 3 1 3 1 3 1 1:1] J.(—ay)‘ dxdz+f f J-(a,)‘ dxdz+f 1.1711,) y=0 2 o y=1 2 o 1(= 3 l =104f f[ cos(2x)e o—cos(2x)e'2] dxdz— 104f f sin(2)e'2ydydz 2 0 l 1 =11)“ (2) 5111(21)‘; (3 ,2)[1, 3—2]+104(§)sin(2)e_2y‘0(3 2 2) = g ldydz 5.3. Let . 400 slné‘ J: r2+4 a, A/n‘r2 a) Find the total current ﬂowing through that portion of the spherical surface I" = 0.8, bounded by 0.17r < 9 < 0.37r, 0 < 4i: < 27:: This willbe 311 400\$i1100(.8)227r 3" . 2 I=ffJ 11‘ «1:1? [1:1 (3)2+ —:.2(8) si11151c11911a11=WI”r 5111 d9 _ —346.5 —[1 — cos(29)]d6— _ 7_7. 4A .11: b) Find the average value of J over the deﬁned area The area is 217 .31? Area =f f (.8)2 sin19d6 d¢ =1.461112 0 .117 The average current density is thus Jaug = (714/146) 11, = 53.0 a, A/mz. 5.5. Let 25 20 = _ _ — A 2 J p :19 p2 + 0.01% /m a) Find the total current crossing the plane 2 = 0.2 in the az direction for p < 0.4: Use I=ffJ-n= da=f2ﬁf4p; —mpdpd¢ = —(§)20]n(.01+p2)‘: (27:): —207rln(l7)= —l78.0A b) Calculate BpU/at: This is found using the equation of continuity: 3p" 1 3 3J3 13 3 ( —20 ) = —v -J= J — = —— 5+ — a: pig—pm pH— 32 paid2 )+ 3—2 p2+.01 =9 c) Find the outward current crossing the closed surface deﬁned by p = 0.01, p = 0.4, 2 = 0, and z = 0.2: This will be 21': 25 2K2 25 1:] [0 _a,. (_ ap)(.01)dqbdz+f [0 _—ap- (apnowdz +foh f0 pr:+.01az'(_aZ)pdpd¢+fozﬂ f0 p2+_01az-(az)pdpd¢—Q since the integrals will cancel each other. d) Show that the divergence theorem is satisﬁed for J and the surface speciﬁed in part b. In part c, the net outward ﬂux was found to be zero, and in part b, the divergence of J was found to be zero (as will be its volume integral). Therefore, the divergence theorem is satisﬁed. 5.9a. Using data tabulated in Appendix C, calculate the required diameter for a 2-m long nichrome wire that will dissipate an average power of450 W when 120 V ms at 60 Hz is applied to it: The required resistance will be R V2 l _ P _ a (1:02) Thus the diameter will be 1P 2(450) _4 V 6an V (106)3(12032 X m b) Calculate the nns current density in the wire: The rms current will be I = 450/120 = 3.75 A. Thus 3 75 J = —2 = 6.0 x 107A/m2 :rr(2.8 x 10-4/2) — 5.11. Two perfectly—conducting cylindrical surfaces are located at p = 3 and p = 5 cm. The total current passing radially outward through the medium between the cylinders is 3 A dc. Assume the cylinders are both oflength I. a) Find the voltage and resistance between the cylinders, and E in the region between the cylinders, if a conducting material having a = 0.05 S/m is present for 3 < p < 5 cm: Given the current, and knowing that it is radially—directed, we ﬁnd the current density by dividing it by the area of a cylinder of radius p and length l : 3 27m! Then the electric ﬁeld is found by dividing this result by a: J = aJo A/m2 3%:9—5531’0“,m E = Zataplap pl The voltage between cylinders is now: . 4. hf... in [925... .pdp_9lii.(g)_ f-_8v Now, the resistance will be v 4. 1. R_T=§=ﬂ _ 31 1 b) Show that integrating the power dissipated per unit volume over the volume gives the total dissipated pow er: We calculate 2.: 32 5 14.64 _ =_ _ =_w I):qu M” f0] [11: (2412p 322—105)!” dp W ‘7“ 24(05)lm(3) I We also ﬁnd the power by taking the product of voltage and current: 4. 14. 4 P=VI=\$(3)=T6W which is in agreement with the power density integration. 5.13. A hollow cylindrical tube with a rectangular cross-section has external dimensions of 0.5 in by l in and a wall thickness of 0.05 in. Assume that the material is brass, for whicho- = 1.5 x 107 81m. A current of200 A dc is flowing down the tube. a) What voltage drop is present across a lm length of the tube? Converting all measurements to meters, the tube resistance over a l m length will be: 1 R‘ = (1.5 x 107) [(2.54)(2.54/2) x 10—4 — 254(1 — .1)(2.54/2)(1 — .2) x 10-4] = 7.38 x 10-4 Q The voltage drop is now v = 1R1 = 20017.38 x 10-4 = 0.147 V. b) Find the voltage drop if the interior of the tube is ﬁlled with a conducting material for which 0' = 1.5 x 105 81m: The resistance ofthe ﬁlling will be: 1 R2 = (1.5 x 1(15)(1/2)(2.54)2 x 10-4(.9)(.8) = 2.87 x10‘2 :2 The total resistance is now the parallel combination of R1 and R2: R7 = R1 Rg/(Rl + R2) = 7.19 x 10'4 Q, and the voltage drop is now V = ZOORT = .144V. 5.15. Let V =10(p + l)22cos¢iV in free space. a) Let the equipotential surface V = 20V deﬁne a conductor surface. Find the equation of the conductor surface: Set the given potential ﬁmction equal to 20, to ﬁnd: (p+ ”22005.? = 2 b) Find p and E at that point on the conductor surface where d: = 0.27: and z = 1.5: At the given values of 0': and 2, we solve the equation of the surface found in part a for p, obtaining p = ﬂ. Then BV 1611/ BV E=—VV=——ap— ——a — az 3p pad: 32: 2 10+] 2 - =—10z cosdiap—l—IO z s1n¢a¢—20(p+l)zcos¢az Then E(.10, 21:, 1.5) = —18.2 ap +145 51¢, — 26.7 aZV/m c) Find |p3| at that point: Since E is at the perfectly-conducting surface, it will be normal to the surface, so we may write: E - E p; = 50E - n = 60— = ens/E - E = Ens/(18.2): + (145)2 + (26.7)2 = 1.32 nC/m2 sm'faoe IEI 5.17. Given the potential ﬁeld lOsz V = V x2 +4 in ﬁ-ee space: a) FindD at the surface 2: = 0: Use 8 x 100.1: E = —VV = —100za—x (x2+4) ax — Day — man/m At a = 0, we use this to ﬁnd 100€ox 2 DLZ = 0) = €0E(Z = 0) = —maz C/m b) Show that the z = 0 surface is an equipotential surface: There are two reasons for this: 1) E at z = 0 is everywhere z-directed, and so moving a charge around on the surface involves doing no work; 2) When evaluating the given potential ﬁmction at z = 0, the result is 0 for all x and y. 0) Assume that the z = 0 surface is a conductor and ﬁnd the total charge on that portion of the conductor deﬁned by 0 < x < 2, —3 < y < 0: We have _ 10060): _ 2 2:0 12 +4 C/m pj=D-az S 0 Q for 100er d”, (3)000) ' 1m 2+4)‘2 150 m2 092 c _3 D x2+4 y 60 2 0 60 n ...
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