sol_hw04_Ch21

sol_hw04_Ch21 - PMsrcs 04- 0M gamma J 21.8. IDENTIFY: Use...

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Unformatted text preview: PMsrcs 04- 0M gamma J 21.8. IDENTIFY: Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum atoms in one sphere. Each atom has 13 electrons. Apply Coulomb's law and calculate the magnitude of charge |q| on each sphere. SET UP: N A = 6.02><1023 atoms/mo] . [q| = nge , where n; is the number of electrons removed from one sphere and added to the other. EXECUTE: (a) The total number of electrons on each sphere equals the number of protons. , 0.0250 k ne =np = (13)(N.)[ g —~-——--— 2 7.25X1024 electrons . 0.026982 kg/mol 1 q2 . . (b) For a force of 1.00x10‘ N to act between the spheres, F =l.00x10" N = —— This gives , . 47150 r‘ |q| = ‘l47rq,(1.00x10‘ N)(0.0800 m)2 = 8.43x104 C . The number of electrons removed from one sphere and added to the other is n; = |q1/e = 5.27x10‘5 electrons. (c)n;/ne=7.27x10"'°. . EVALUATE: When ordinary objects receive a net charge the fractional change in the total number of electrons in g the object is very small . . 21.33. IDENTIFY: Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between the plates. The force is constant and produces a constant acceleration. The motion is similar to projectile motion; use constant acceleration equations for the horizontal and vertical components of the motion. (3) SET UP: The motion is sketched in Figure 21.33a. 2.00 cm 0.50 cmi v I ’ For an electron q = —e. oi» - — E Figure 21.33a F = qi and q negative gives that 17‘ and E are in opposite directions, so 17‘ is upward. The free-body diagram i for the electron is given in Figure 21.33b. y 1 ‘ EXECUTE: 217‘ =ma} d F=€E eE=ma I Figure 21.33b Solve the kinematics to find the acceleration of the electron: Just misses upper plate says that x - x0 = 2.00 cm when y — y0 = +0500 cm. x-component v0} = v0 =l.60x10° m/s, a\ =0. x—-x0 = 0.0200 m, t= ? x— xH = vott + gaxt‘ 21.51. z= “"0 =———°'0202 m =1.25><10‘3 s vm l.60><10 m/s In this same time t the electron travels 0.0050 m vertically: y-component [-1 1.25X10‘857 VOy 20‘ y- Yo m’ “1:? y — yo = vmt + ilayt2 a = 202—») =W=640x10n m/s2 " t2 (1.25x10'“ s)‘ (This analysis is very similar to that used in Chapter 3 for projectile motion, except that here the acceleration is upward rather than downward.) This acceleration must be produced by the electric-field force: eE = ma e l.602><10"9 C Note that the acceleration produced by the electric field is much larger than g, the acceleration produced by gravity, so it is perfectly ok to neglect the gravity force on the elctron in this problem. (b) a--————————_27———-=3.49x10'° m/s2 m 1.673X10 kg This is much less than the acceleration of the electron in part (a) so the vertical deflection is less and the proton won’t hit the plates. The proton has the same initial speed, so the proton takes the same time t = l.25><10'fl s to travel horizontally the length of the plates. The force on the proton is downward (in the same direction as E, since q is positive). so the acceleration is downward and av = —3.49><10'° m/sz. y— yo = voytifiavt2 =§(—3.49x10'° rn/s2)(l.25><10'B s)2 = —2.73><104’ m. The displacement is 2.73><10'6 m, downward. (c) EVALUATE: The displacements are in opposite directions because the electron has negative charge and the proton has positive charge. The electron and proton have the same magnitude of charge, so the force the electric field exerts has the same magnitude for each charge. But the proton has a mass larger by a factor of 1836 so its acceleration and its vertical displacement are smaller by this factor. IDENTIFY: The resultant electric field is the vector sum of the field E] of ql and El of qz. ‘ SET UP: The placement of the charges is shown in Figure 21.5 la. .V ' a <———> <—> h 0.150m 0.150m 0.150m 0400711 .C Figure 21.513 EXECUTE: (a) The directions of the two fields are shown in Figure 215”). y E] = E2 = 471350 r‘ E] .9 o _ 1 C a x E, = (8.988x109 N . m‘/C2).6_9.L2 ‘12 < 0 £2 q > 0 m) 1 E1 = E2 = 2397 N/C Figure 21.51b E” = —2397 N/C, EU, :0 Ex = —2397 N/C, E2). = 0 EX = E” + E2x = 2(—2397 N/C) = —4790 N/C Ey = E“. + E2), = 0 The resultant electric field at point a in the sketch has magnitude 4790 N/C and is in the —x-direction. ; i (b) The directions of the two fields are shOWn in Figure 21.51e. y 52 51 <———? q2 < 0 q1 > 0 b Figure 21.51e —9 E1 =—1—li;-| = (8.988x109 N .m2/C2)6—'°21‘1—0—2C- = 2397 N/C 47:60 rl (0.150 m) .9 E2 = LEE—1 = (8.988x109 N - mZ/CZ)M = 266 N/C 47:60 r2 (0.450 m) E” = +2397 N/C, Eh. = 0 E21 = -266 N/C, E2). = 0 Ex = E” + Eh = +2397 N/C — 266 N/C = +2130 N/C E), = E”. + E2), = 0 The resultant electric field at point b in the sketch has magnitude 2130 MC and is in the +x—direction. (c) The placement of the charges is shown in Figure 21 .51d. 0300 m sin0 = 0300 m = 0.600 0.500 m c050 = 0400 m = 0.800 0.500 m Figure 21.51d The directions of the two fields are shown in Figure 21 .51e. y E] =-1_|_q;_| q2<0 ql>ox 47:60 r1~ ‘ \ 6.00x10'9 C E, =(8.988x10° N~m2/C2) \ \ my 5;" ’ ‘ ‘ E. =337.0 N/C =__1 I221 47560 r22 .9 E =(8-988x10" N ~ Maw - (0.500 m) E, = 215.7 N/C Figure 21.51e E” = 0, El}. _ —E1 = —337.0 N/C — Eh = —E2 sin0 = —(215.7 N/C)(0.600) = —129.4 N/C EN = +E2 c050 = +(215.7 N/C)(O.800) = +172.6 N/C E‘ =5“ +63 =—129 N/C = EH +15% =-337.0 N/C +172.6 N/C = —164 N/C EzJE‘f +53 =209 N/C The field E and its components are shown in Figure 21.51f. X E tana = —" 1 —164 N/C _ —129 N/C a = 232°. counterclockwise from + x—axis tuna: +1271 Figure 21.51f (d) The placement of the charges is shown in Figure 21 .51 g. d 0.200 In / X sin0: 7‘ =0.800 0.250 m, " \ \0150 m (51.5%m / \ cost? = ' m = 0,600 09‘ ‘1! 0.250 m ().l50m 0.150m X Figure 21.51g The directions of the two fields are shown in Figure 21.51h. E. : E1 = 41:60 r‘ } ,9 E1 = (8.988x10° waaw (0.250 m)‘ E = 862.8 N/C E2 = E, = 862.8 N/C Figure 21.51h 15, = E.. + E;. = ~2(862.8 N/C)(0.600) = —1040 N/C E13=+El8in9' E2), =_E: Sin9 E. =E,\ +E2‘, :0 E = 1040 N/C, in the — x-direction. EV’ALUATE: The electric field produced by a charge is toward a negative charge and away from a positive charge. As in Exercise 21.45, we can use this rule to deduce the direction of the resultant field at each point before doing any calculations. } Elr = —El cosé), E1 = ~E3 c080 21.53. IDENTIFY: Apply Eq.(21.10) for the finite line of charge and E = L for the infinite line of charge. 1 272'60 . SET UP: For the infinite line of positive charge, E is in the +x direction. i EXECUTE: (a) For a line of charge of length 2a Centered at the origin and lying along the y-axis, the electric field 1 A A l - l i is given by Eq.(21.10): E = ————-———1 . 27’50 xtlxz/az +1 . . . . — x1 9 (b) For an infinite line of charge: E = 2”qu l . Graphs of electric field versus position for both distributions of charge are shown in Figure 21.53. ' EVALUATE: For Srnall x, close to the line of charge, the field due to the finite line approaches that of the infinite lilnefof charge. As x increases, the field due to the infinite line falls off more slowly and is larger than the field of t e mite ine. Top: infinite Bottom: finite : 0.00 1.00 2.00 3.00 I 4.00 (%) Figure 21.53 21.55. 21.56. 21.58. IDENTIFY: For a ring of charge, the electric field is given by Eq. (21.8). E = qE . In part (b) use Newton's third l law to relate the force on the ring to the force exerted by the ring. ‘ SET UP: Q = 0.125x10'° C, a = 0.025 m and x = 0.400 m. EXECUTE: (a) E‘ =—1———,—Qx,—;/—,f=(7.0 N/Cif. 47ml (x‘ +a‘)’ ‘ (b) Fm.“ = —F‘mq =—qE = —(—2.50x10*’ci (7.0 N/Cii = (1.75x10—5 N)i EVALUATE: Charges q and Q have opposite sign, so the force that q exerts on the ring is attractive. IDENTIFY: We must use the appropriate electric field formula: a uniform disk in (a), a ring in t b) because all the charge is along the rim of the disk. and a point-charge in (c). (3) SET UP: First find the surface charge density (Q/A), then use the formula for the field due to a disk of charge. E _gil__;_i ‘ 26% JtR/xf +1 1 6.50 10‘”C J , EXECUTE: The surface charge density is a =9=£ = = 1324 x 10 ' C/m‘. A ltr‘ Ir(0.0125 m)“ The electric field is 1,324x10'5 C/m2 2t8.85><10‘” CZ/N‘mlt E 1_____l__. ‘260 JtR/xthri E‘ = 1.14 X 105 N/C, toward the center of the disk. . l V (b) SET UP: For a ring of charge, the field 18 E = . 4m (.r' + a") EXECUTE: Substituting into the electric field formula gives 1 Qt E _ ##W _ 19.00x10” Nrm1/C11(6.50><10’” C)(0.0200 m) 4’”. (xl +031 [100200 m)1+(0.0125 me” E = 8.92 x 10‘ MC, toward the center of the disk. (c) SET UP: For a point charge, E = (l/4Iz'e,)q/ rll EXECUTE: E = (9.00 x 109 N- m3/C31650 x 10 '9 ci/(0.0200 mf = 1.46 x 105 N/C (d) EVALUATE: With the ring. more of the charge is farther from P than with the disk. Also with the ring the component of the electric field parallel to the plane of the ring is greater than with the disk. and this component cancels. With the point charge in (c), all the field vectors add with no cancellation. and all the charge is closer to point P than in the other two cases. IDENTIFY and SET UP: The electric field produced by an infinite sheet of charge with charge density O'has magnitude E = 13—i— . The field 15 directed toward the sheet if it has negative charge and is away from the sheet if it ~50 has positive charge. EXECUTE: (a) The field lines are sketched in Figure 21.58a. (b) The field lines are sketched in Figure 21.58b. EV’ALUATE: The spacing of the field lines indicates the strength of the field. in part (a) the two fields add between the sheets and subtract in the regions to the left of A and to the right of B. In part (b) the opposite is true. A B A B Hr ~20 +0- + 20' (a) (b) Figure 21.58 21.67. IDENTIFY: Like charges repel and unlike charges attract. The force increases as the distance between the charges decreases. '1 SET UP: The forces on the dipole that is between the slanted dipoles are sketched in Figure 21.6721. l. I EXECUTE: The forces are attractive because the + and — charges of the two dipoles are closest. The forces are _ toward the slanted dipoles so have a net upward component. In Figure 21 .67b, adjacent dipoles charges of opp051te sign are closer than charges of the same sign so the attractive forces are larger than the repulsive forces and the dipoles attract. . I EVALUATE: Each dipole has zero net charge, but because of the charge separatlon there 15 a non-zero force between dipoles. Fl F U Fly F2 . Fix ED 1 EDGE) (a) (b) Figure 21.67 21.74. IDENTIFY: Apply 2F: = O and ZR, = 0 to one of the spheres. SET UP: The free-body diagram is sketched in Figure 21.74. Fe is the repulsive Coulomb force between the spheres. For small 19, sine : tan (9 . Exacum 212 =Tsin6—Ifl =0and 2F. =Tcost9—mg :0. So mgsmg=iz =k—‘f. But tanazsln0=i c0319 2L ’ 2 1 1/3 so (13:2k" Land d: ‘1 L l . mg Zirsflmg ) EVALUATE: (1 increases when q increases. T F. T sin 0 : mg Figure 21.74 21.90. IDENTIFY: Use Eq. (21.7) to calculate the electric field due to a small slice of the line of charge and integrate as in Example 21.11. Use Eq. (21.3) to calculate F. SET UP: The electric field due to an infinitesimal segment of the line of charge is sketched in Figure 21.90. Figure 21.90 Slice the charge distribution up into small pieces of length dy. The charge dQ in each slice is dQ = Q(dyla). The electric field this produces at a distance x along the x-axis is dE. Calculate the components of dB“ and then integrate over the charge distribution to find the components of the total field. 21 .93. d dE=——l— fig»): Q (fly a] 4/reU x‘+_v‘ 472'6011 x +.V d. (15“ =dEcos€= Qx 475.61%) (X‘ + y’) d. dEV=—dEsin0=— Q ‘ 47150“ (x‘+y') EXECUTE: E—dE- Qx JUL: 9" i__Y_u= 9-1—— *—.l. "~ 47r50a 0(x:+y3)'v2 47tena x2.lx2+y1 0 471'e,p\"/;1¢2+a2 “ l _ _ Q .. ydy =_Q[__1_=‘__Q_[l____J Ey—J‘dEy— (mew [0(x2+y2)312 4fleoal— 0 47:60a x x2+a2 (b)F“=q0E -qQ 1 . - 4Q 1_; F=—E= ~—-—.F.=—qE.— — . . I q ‘ 4qu,x./x3+a3 ' 4/reoa x x-+a- 1 3"” 1 a2’ 1 a2 (c)Forx>>a,—_1_=_(1+fl?) =_(1-__J=_fl__3. x2+a2 X\ x x 2x2 x 2x 1 1 13 a F,“ qQ ’F : qQ [_‘_+¢_3_)= £19 3 " 4Ir6012 4lreoa x x 2): Site“): EVALUATE: For x >> a, F\. << F‘ and F =|Fr| = 4’3Q 2 and I7 is in the —x—direction. Forx >> a the charge qrr distribution Q acts like a point charge. IDENTIFY: Apply Eq.(21.11). SET UP: d=Q/A=Q/7tR2. (1+ yly'” =1—y3/2 . when y2 «1. EXECUTE: (a) E=2£[l—(R2/x2+l)_w]. 60 2 2 "/2 E=W 1~ (“ism—1+1 =0.s9 N/C,in the +x direction. 2st, (0.200 m) 2 2 (b)For x>>R 5:1[1—(1—R3/2x2+.~)]=£~L= “R2 = Q ,, 260 260 2x“ 47tsox 47r60x' (c) The electric field of (a) is less than that of the point charge (0.90 N/C) since the first correction term to the point charge result is negative. (0.90—0.89) =0.01=1% . For x=0.100m, 0.89 (d) For x = 0.200 m , the percent difference is Ed“k = 3.43 N / C and Ewm = 3.60 N / C , so the percent difference is = 0.047 = 5%. EVALUATE: The field of a disk becomes closer to the field of a point charge as the distance from the disk increases. At x = 10.0 cm, R / x = 25% and the percent difference between the field of the disk and the field of a point charge is 5%. 21.95. IDENTIFY: Find the resultant electric field due to the two point charges. Then use 17‘ = qE' to calculate the force 21.99. 21.104. IDENTIFY: on the point charge. k SET UP: Use the results of Problems 21.90 and 21.89. ' . I EXECUTE: (a) The y-components of the electric field cancel, and the x-component from both charges, as given in 1—2Q1 1 ~ 1—2qu 1 j: ' =——-— —————— .Therefore,F=—-————— ——-——-—- l .Ify>>a Problem 2190,15 E. 471% a [y (y: + a2)1(2] 41:60 a y (yz +a2)1(2 — — . . c 1 i t F z—l— ZQq (l—(l—a‘l2y‘ +~-))t =———Qq‘al. 47:60 ay 471's0 y‘ (b) If the point charge is now on the x-axis the two halves of the charge distribution provide different forces, -1_2‘l( 1 -1); though still along the x-axis, as given in Problem 21.89: I?+ = ql7I+ = 4m0 a x—a x .. - e — ~ — l 1 2 l r and F =qE_=——1—g(—l—— 1 )i .Therefore, F=F++FV=__g _._.__+ ), _For x>>a, — 4fl60 a x x+a 47:60 a x—a x x+a ~ 3 2 2 1 2 a? xiii HEM—2+... -2+[1_£ta_._...] .=__ 9:1 .. 472's0 ax x x x x" 47:50 x EVALUATE: If the charge distributed along the x-axis were all positive or all negative, the force would be proportional to 1/ yz in part (a) and to 1/ x2 in part (b), when y or x is very large. IDENTIFY: Each wire produces an electric field at P due to a finite wire. These fields add by vector addition. c . 1 Q Each held has magnitude ~——————~———7 47’50 mix" +a‘ the field due to the positive wire points downward, making the two fields perpendicular to each other and of equal magnitude. The net field is the vector sum of these two, which is Enet = 2E, cos 45‘ = 2; Q c0545O . In SET UP: . The field due to the negative wire points to the left, while part (b). the electrical force on an electron at P is cE. Exrcmr: (a) The net field is = Z—LLCOMSO . 4fl'éo xx/x2 + a2 9 2(9.00><109N - mz/C:)(2.50><10"’C)cos45° (0.600 m) (0.600 m)2 + (0.600 m): The direction is 225° counterclockwise from an axis pointing to the right through the positive wire. Ens = = 6.25 x 10‘ N/C. (b) F = eE = (1.60 x 10‘[9 C)(6.25 x 104 MC) = 1.00 x 10‘H N, opposite to the direction of the electric field, since the electron has negative charge. EVALUATE: Since the electric fields due to the two wires have equal magnitudes and are perpendicular to each other. we only have to calculate one of them in the solution. Apply Eq.( 21.1 1) for the electric field of a disk. The hole can be described by adding a disk of charge density -0' and radius Rl to a solid disk of charge density +0' and radius R: . SET UP: The area of the annulus is 7t(R22 — Rl2)0' . The electric field of a disk, Eq.(21.11) is hip—MW]. EXECUTE: (a) Q = A0 = 7r(R22 — Rho (b) E(x)=—0—([1—1/ (Rz/x)2+1]—[1—1/\/m])lilf. E(x)=§(1/W—1/m)lflf. x 260 x The electric field is in the +x direction at points above the disk and in the —x direction at points below the disk, and lxl the factor —i specifies these directions. x 2 (c) Note that 1/,/(R,/x)2+1=m(1+(x/R,)2y“Z am. This gives E(x)=l i—i £121 i—i xi. R, R, 2% R, R2 Sufficiently close means that (x / R,)2 << 1. (d) F, = qE, =-—-[7:--—l%-]x . The force is in the form of Hooke’s law: F, = —kx, with k =—{—-——]. l 2 f__1_ 5-; 11L_L 22: m 2;: 2a,»: R, R2' EVALUATE: The frequency is independent of the initial position of the particle, so long as this position is sufficiently close to the center of the annulus for (x/ RI )2 to be small. 21.107. IDENTIFY: To find the electric field due to the second rod. divide that rod into infinitesimal segments of length I dx. calculate the field dE due to each segment and integrate over the length of the rod to find the total field due to i the rod. Use dfi = dq E to find the force the electric field of the second rod exerts on each infinitesimal segment of thefirstrod. I SET UP: An infinitesimal segment of the second rod is sketched in Figure 21.107. dQ=(Q/L)dx'. 2 EXECUTE: (a) dE=————k‘Q——,—,-=k—Q————d‘5—fi. (x+a/2+L—x)' L (x+a/2+L—x)‘ E 4215 _Q 1 ‘ 0 ‘ L 0(x+a/2+L~x’)2 L x+a/2+L-x’,, L ,x+a/2 x+a/2+L,. ‘ L 2x+a_2L+2x+a (b) Now consider the force that the field of the second rod exerts on an infinitesimal segment dq of the first rod. This force is in the +x—direction. dF = dq E . I f I f “252,1 #1 j. E , i l F: IE dq: IBM/2Q dx=2k?- IBM/T l —————-——-——l )dx. "/2 L L* "/3 2x+a 2L+2x+a p 7 2 ’1 '1 2 / 17—ka l([1n(a+2x)]5;‘1‘-—[1n<2L+2x+a)]§*:")=kg 1n (———“+2L+“][2L+2") . L“ 2 ' " L 2a 4L+2a 2 2 ‘ 1n 1%.. . L‘ a(a+2L) 2 2 / 2 2 (c)For a>>L. F=kQ, ln(-a—,£]—t£’—a)— =£Q—,—(21n(1+L/a)—ln(l+2L/a)). ; L” a”(l+2L/a) L‘ i Forsmallz, ln(l+z)z:—E—.Therefore. for a>>L, sz—q 2 £— L~2+w — ill—21:, +--- sz,-. 2 L‘ a 2a a a' a’ ; EVALUATE: The distance between adjacent ends of the rods is a. When a >> L the distance between the rods is much greater than their lengths and they interact as point charges. Figure 21.107 ...
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This note was uploaded on 04/08/2011 for the course PHYSICS 1B taught by Professor Corbin during the Spring '11 term at UCLA.

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sol_hw04_Ch21 - PMsrcs 04- 0M gamma J 21.8. IDENTIFY: Use...

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