f5ch23 - 636 CHAPTER 23 ELECTRIC FIELDS CHAPTER 23 Answer...

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Unformatted text preview: 636 CHAPTER 23 ELECTRIC FIELDS CHAPTER 23 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (a) rightward; (b) leftward; (c) leftward; (d) rightward; (p and e have same charge magnitude and p is farther) all tie (a) toward positive y; (b) toward positive x; (c) toward negative y (a) leftward; (b) leftward; (c) decrease (a) leftward; (b) 1 and 3 tie, then 2 and 4 tie Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. (a) toward positive x; (b) downward and to the right; (c) A (a) to their left; (b) no two points: one to the left of the particles, the other between the protons q=40 d2 , leftward (a) yes; (b) toward; (c) no (the eld vectors are not along the same line; (d) cancel; (e) add; (f) adding components; (g) toward negative y all tie (a) 3, then 1 and 2 tie (zero); (b) all tie; (c) 1 and 2 tie, then 3 e, b, then a and c tie, then d (zero) (a) rightward; (b) +q1 and q3 , increase; q2 , decrease; n, same (a) toward the bottom; (b) 2 and 4 toward the bottom, 3 toward the top a, b, c (a) positive; (b) same (a) 4, 3, 1, 2; (b) 3, then 1 and 4 tie, then 2 CHAPTER 23 ELECTRIC FIELDS 637 14. Accumulated excess charge creates an electric eld; proximity of a second body con centrates the excess charge and increases the eld, causing electrical breakdown in the air (sparking). Solutions to Exercises & Problems 1E (a) FA = eEA = (1:60 10 (b) EB = EA =2 = 20 N/C. 2E 19 C)(40 N/C) = 6:4 10 18 C: +q The lines of forces due to +Q and Q are similar to those in Fig. 23-5. Just turn the book by 90 so that the line joining the two charges in Fig. 23-5 becomes horizontal, with +Q on the left. Obviously the force on +q is to the right. 4E 3E q1 -q 2 q1 -q2 q1 -q 2 q1 > q 2 q1 = q2 q 1 < q2 638 CHAPTER 23 ELECTRIC FIELDS 5E The diagram to the right is an edge view of the disk and shows the eld lines above it. Near the disk the lines are perpendicular to the surface and since the disk is uniformly charged the lines are uniformly distributed over the surface. Far away from the disk the lines are like those of a single point charge (the charge on the disk). Extended back to the disk (along the dotted lines of the diagram) they intersect at the center of the disk. If the disk is positively charged the lines are directed outward from the disk. If the disk is negatively charged they are directed inward toward the disk. Lines below the disk are exactly like those above. 6E Solve q from E = q=40 r2 : q = 40 Er2 = (1:00 N/C)(1:00 m)2 = 1:11 10 8:99 109 N m2 =C2 10 C : 7E Since the magnitude of the electric eld produced by a point charge q is given by E = q=40 r2 , where r is the distance from the charge to the point where the eld has magnitude E , the charge is 2 N/C) q = 40 r2 E = 8(0:50 m) 9(2:0m2 /C2 = 5:6 10 :99 10 N 11 C : 8E 2Q = 8Q = (8(2:0 10 7 C)(8:99 109 N m2 =C2 ) = 6:4 105 N/C : E = 4 (r=2)2 4 r2 (0:15 m)2 0 0 points toward the negative charge. (b) F = eE = (1:60 10 19 C)(6:4 105 N/C) = 1:0 10 charge. E 9E 13 N: F points toward the positive Since the charge is uniformly distributed throughout a sphere the electric eld at the =0 e equ q q1 + Set the net forc 0 (x0 x)2 Fnet = F + F = 40x2 40 es in th ) = 6:0 mm.d by either of the two chargve is direc 0 0 = 3x = 3(2:0 mm ce ati x Solve for x : x n of the electric eld produ net eld is also in the neg (b) The directio = 2:0 mm. Therefore the x direction at x 11P d x curve is plotted The E (x) vs. x E(x) = d=( above. Here x0 p 3 + 1), where E = 0 x0 640 CHAPTER 23 ELECTRIC FIELDS (a) Convince yourself that a point where E = 0 can only be located to the right of the positive charge. Denote the separation between the point and the positive charge by x, then 5:0q = 0 : 1 2:0q E = 4 x2 (x + d)2 0 Solve for x: x = 1:7d. (b) 12P (a) E 13P A points to the left. 1 : EA = 4 1d0q 2 0 2:0q = q : (2d)2 80 d2 1 1: 2: EB = 4 (d=0q 2 + (d=0q 2 = 3qd2 : 2) 2) 0 0 E B points to the right. 1 : EC = 4 2d0q 2 0 E 1:0q = 7q : (2d)2 160 d2 (b) C points to the left. +q CHAPTER 23 ELECTRIC FIELDS 641 At points between the charges the individual electric elds are in the same direction and do not cancel. d Charge q2 has a greater magnitude than charge q1 so q2 q1 P a point of zero eld must be closer to q1 than to q2 . It must be to the right of q1 on the diagram. Put the origin at q2 and let x be the coordinate of P , the point where the eld vanishes. Then the total electric eld at P is given by 1 q E = 4 x2 0 2 14P (x d)2 ; q1 where q1 and q2 are the magnitudes of the charges. If the eld is to vanish, q2 =x2 = q1 =(x d)2 . Take the square root of both sides to obtain pq1 =x = pq2 =(x d). The solution for x is pq2 p4:0q x = pq pq d = p4:0q 1pq d 2 1 1 1 = 2:02:01:0 d = 2:0d = (2:0)(50 cm) = 100 cm : The point is 50 cm to the right of q1 . 15P (a) The elds from the two charges of 5:0q each cancel out. So 1 : Ep = 4 3d0q 2 0 16P 12q = 0 : (2d)2 Denote the electron with subscript e and the proton with p. From the gure to the right we see that the jEe j = jEp j = e=(40 d2 ), where d = 2:0 10 6 m, and the magnitude of the net electric eld is x Ep E y E = 2Ee cos = 2 4e d2 cos 0 19 C)(8:99 109 N m2 =C2 ) (cos 60 ) = 2(1:6 10 2:0 10 6 m2 = 3:6 102 N/C : Ee = 60 proton electron 642 CHAPTER 23 ELECTRIC FIELDS In the gure to the right, each of the six arrows represents the net electric eld of a pair of diametrically opposite charges (e.g. the one which points vertically upward represents the eld of the 12q and the 6q charges). Since each of the six E- elds has a magnitude of E = 6q=40 r2 , where r is roughly the radius of the clock, by symmetry the net E- eld should point at A, i.e., the 9:30 mark. 17P 11 10 A 9 8 7 12 1 2 3 4 5 6 By symmetry the elds due to the two electrons placed in line with the midpoint of a side cancel out. So 1 E = 4 p e 2 = 3e a2 0 ( 3a=2) 0 19 C)(8:99 109 N m2 =C2 ) = 4(1:60 10 3(0:20 m)2 = 4:8 10 8 N/C : E is perpendicular to the side line and points toward the corner of the triangle opposite to the side. 19P 18P By symmetry the elds due to the two charges of magnitude q each cancel out. So 0q 1 2:p i = q i ; Ep = 40 (a= 2)2 0 a2 where i is a unit vector from the +2:0q charge to point P . Put the origin of a coordinate system at the center of the triangle, where the net electric eld is given by E = E1 + E2 + E3 : From symmetry consideration it is obvious that E1 = E2 = E3 . This gives Q = q1 = q2 = +1:0 C. q1 =1.0 C x q3=Q 20P E2 E1 y E3 q2=1.0 C CHAPTER 23 ELECTRIC FIELDS 643 From symmetry, the only two pairs of charges which produce a non-vanishing eld E are: pair 1, which is the one in the middle of the two vertical sides of the square (the +q, 2q pair); and pair 2, the +5q, 5q pair. Denote the electric elds produced by each pair as E1 ans E2 , respectively. Set up a coordinate system as shown to the right, with the origin at the center of the square. Now 1 q + 2q = 3q E1 = 4 d2 d2 40 d2 0 and 1 5 E2 = 4 p5q 2 + p5q 2 = 4qd2 ; ( 2d) 0 ( 2d) 0 21P y -5q E2 E x E1 5q -q +q so the components of the net E eld are given by Ex = E1x + E2x = E1 + E2 cos 45 3q + 5q cos 45 = 6:536 q = 4 d2 40 d2 40 d2 0 and Ey = E1y + E2y = E2 sin 45 = q p Thus the magnitude of E is 40 d2 5q sin 45 = 3:536 q 40 d2 : q 2 2 E = Ex + Ey = (6:536)2 + (3:536)2 4q d2 = 47:43d2 ; 0 0 and E makes an angle with the positive x axis, where = tan 1 Ey = tan Ex 1 3:536 = 28:4 : 6:536 644 CHAPTER 23 ELECTRIC FIELDS Choose the coordinate axes as shown on the diagram to the right. At the center of the square the electric elds produced by the charges at the lower left and upper right corners are both along the x axis and each points away from the center and toward the charge that produces it. Since each charge is a p p distance d = 2a=2 = a= 2 away from the center the net eld due to these two charges is 22P y q d a x -2q d -q 2q 1 q Ex = 4 a2=2 a2q=2 2 0 1 q = (8:99 109 N m2 /C2 )(1:0 10 8 C) = 7:19 104 N/C : = 4 a2 =2 (0:050 m)2 =2 0 At the center of the square the eld produced by the charges at the upper left and lower right corners are both along the y axis and each points toward the center, away from the charge that produces it. The net eld produced at the center by these charges is 1 q Ey = 4 a2=2 2 0 q 1 q 4 2 =2 = 40 a2 =2 = 7:19 10 N/C : a p The magnitude of the eld is 2 2 E = Ex + Ey = 2(7:19 104 N/C) = 1:02 105 N/C q and the angle it makes with the x axis is = tan 1 Ey = tan 1 (1) = 45 : Ex It is upward in the diagram, from the center of the square toward the center of the upper side. The magnitude of the dipole moment is given by p = qd, where q is the positive charge in the dipole and d is the separation of the charges. For the dipole described in the problem p = (1:60 10 19 C)(4:30 10 9 m) = 6:88 10 28 C m. The dipole moment is a vector that points from the negative toward the positive charge. 23E CHAPTER 23 ELECTRIC FIELDS 645 1 E = 4 (z q 2)2 + (z + q 2)2 : d= d= 0 For z d we have (z d=2) 2 z 2 , so 1 q + q = 2q : E 4 z2 z2 40 z 2 0 25E 24E y +q d/2 r P x E2 d/2 -q E1 E y direction. Its magnitude is From the gure above it is obvious that the net electric eld at point P is in the negative 1 E = 2E1 sin = 2 4 (d=2)q + r2 2 0 qd 1 = 4 2 + r2 ]3=2 : 0 [(d=2) p d=2 (d=2)2 + r2 For r d [(d=2)2 + r2 ]3=2 r3 so the expression above reduces to 1 E 4 qd : 0 r3 Since p = (qd)( j), 1 p: E 40 r3 26P Think of the quadrupole as composed of two dipoles, each with a dipole moment of magnitude p = qd. The two dipoles point in opposite directions and produce elds in opposite 646 CHAPTER 23 ELECTRIC FIELDS directions at points on the dipole axis. Consider a point on the axis a distance z above the center of the quadrupole and take an upward pointing eld to be positive. Then the eld produced by the upper dipole of the pair is qd=20 (z d=2)3 and the eld produced by the lower is qd=20 (z + d=2)3 . Use the binomial expansions (z d=2) 3 z 3 3z 4 ( d=2) and (z + d=2) 3 z 3 3z 4 (d=2) to obtain E = 2qd z13 + 23zd4 0 1 + 3d = 6qd2 : z3 2z4 40 z 4 Let Q = 2qd2 , then 3 E = 4Qz4 : 0 27E Use Eq. 23-16. Take E > 0 for E pointing in the positive z direction. The plot is shown below. 4 1 10 5000 E (N/C) 0 5000 1 10 4 10 5 0 5 10 z (cm) 28E From Eq. 23-16 the magnitude of the eld at point P produced by ring 1 is given by R E1 = 4 (Rq1+ R2 )3=2 ; 2 0 CHAPTER 23 ELECTRIC FIELDS 647 while that by ring 2 is Since E1 and E2 have opposite directions the condition for E1 + E2 = 0 is E1 = E2 , or 2 E2 = 4 [(2qR(2R) R2 ]3=2 : )2 + 0 q1 R 2 + R2 )3=2 (R = q2 (2R) : [(2R)2 + R2 ]3=2 Solve for q1 =q2 to obtain q1 =q2 = 2(2=5)3=2 0:51. Set dE=dz = 0 in Eq. 23-16: 2 2 )3=2 2 2 1=2 qz dE = d = q(z + R4 (z 2 3qz (z)3+ R ) = 0 ; dz dz 40 (z2 + R2 )3=2 0 + R2 29P i.e., (z 2 + R2 )1=2 3z = 0. Solve for z : z = R= 2. The electric eld at a point on the axis of a uniformly charged ring, a distance z from the ring center, is given by E = 4 (z2qz R2 )3=2 ; 0 + where q is the charge on the ring and R is the radius of the ring (see Eq. 23-16). For q positive the eld points upward at points above the ring and downward at points below the ring. Take the positive direction to be upward. Then the force acting on an electron on the axis is F = 4 (z2eqz R2 )3=2 : 0 + For small amplitude oscillations z R and z can be neglected in the denominator. Thus 30P p eqz F = 4 R3 : 0 The force is a restoring force: it pulls the electron toward the equilibrium point z = 0. Furthermore, the magnitude of the force is proportional to z , just as if the electron were attached to a spring with spring constant k = eq=40 R3 . The electron moves in simple harmonic motion with an angular frequency given by k = r eq ; != m 40 mR3 where m is the mass of the electron. r 648 CHAPTER 23 ELECTRIC FIELDS By symmetry, each of the two rods produces the same electric eld E0 pointing in the +y axis at the center of the circle. So the net eld is E 31P = 2E0 j = 2j Z = 4 R2 0 1 4q cos dq = 2j Z +=2 cos qRd 40 R2 =2 40 R2 R j : By symmetry, both +q and q produce an equal value of electric eld, E0 , pointing vertically downward. The magnitude of EP is thus 32P EP = 2E0 = 2 Z rod sin dq = 2 Z =2 qr sin d = q ; 40 r2 40 r2 (r=2) 2 0 r2 0 and EP points vertically downward. By symmetry we only need to consider the ycomponent of EP . Consider an in nitesimal segment of length dx in the rod. We have 33P qdx=L dEy = cos dE = y 4 r2 = q4cos d ; r 0 Ly 0 where we used x = y tan and r = y= cos . Thus m q cos d dEy P m dE 40 Ly = 40 Ly sin m qL=2 p = x 20 Ly y2 + (L=2)2 O dx L/ 2 q p = : 20 y 4y2 + L2 Note that the factor of 2 in the rst step above is due to the fact that each half of the rod contributes equally to E . E=2 Z 2q r y 0 CHAPTER 23 ELECTRIC FIELDS 649 (a) = q=L. (b) E 34P P = 4 0 0 i Z L qi Z L q( i) dx dx 2 = 40 L 0 (a + L x)2 = 40 a(L + a) : (a + L x) ( 4q ia)2 ; 0 (c) If a L then a(L + a) a2 , and E P indeed the electric eld of a point charge. Consider an in nitesimal section of the rod of length dx, a distance x from the left end, as shown in the diagram to the right. It contains charge dq = dx and is a distance r from P . The magnitude of the eld it produces at P is given by 1 dE = 4 rdx . 0 2 The x component is 1 dEx = 4 rdx sin 0 2 and the y component is 1 dEy = 4 rdx cos . 0 2 35P y dq x x R r P dE Use as the variable of integration. Substitute r = R= cos , x = R tan , and dx = (R= cos2 ) d. The limits of integration are 0 and =2 rad. Thus Z =2 sin d = cos =2 = : Ex = 4 0 40 40 R 0 0 Z =2 cos d = sin =2 = : Ey = 4 0 40 40 R 0 0 Notice that Ex = Ey no matter what the value of R. Thus E makes an angle of 45 with the negative x axis for all values of R. and 650 CHAPTER 23 ELECTRIC FIELDS 36E From Eq. 23-24 E = 2 1 p 2 z 2 0 z +R 5:3 C/m2 = 2(8:85 10 12 C2=N m2 ) 1 37P 12 cm 3 p 2 + (2:5 cm)2 = 6:3 10 N/C : (12 cm) (a) Let E = =20 = E0 = 3 106 N/C. Thus 10 m) (3: q = R2 = 20 R2 E0 = (2:5 2(8:99 109 N0m210 2N/C) = 1:0 10 7 C : =C ) 2 2 6 (b) (c) The fraction is 10 2 2 N = 0:(2:5 10 18m)2 = 1:3 1017 : 015 m 19 C) 7 10 q frac = Ne = (1:3 10:17 10 C )(1:6 10 = 5:0 10 6 : 38P At a point on the axis of a uniformly charged disk a distance z above the center of the disk the magnitude of the electric eld is E = 2 1 p 2 z 2 ; 0 z +R where R is the radius of the disk and is the surface charge density on the disk. The magnitude of the eld at the center of the disk (z = 0) is Ec = =20 . You want to solve for the value of z such that E=Ec = 1=2. This means 1 1 p 2z 2 = 2 z +R or z + R2 Square both sides, then multiply them by z 2 + R2 to obtain z 2 = (z 2 =4) + (R2 =4). Thus p z2 = R2 =3 and z = R= 3. p 2z 1 = 2: CHAPTER 23 ELECTRIC FIELDS 651 39E The magnitude of the force acting on the electron is F = eE , where E is the magnitude of the electric eld at its location. The acceleration of the electron is given by Newton's second law: 19 C)(2 4 F eE a = m = m = (1:60 10:11 10 :00 10 N/C) = 3:51 1015 m/s2 : 31 kg 9 e e 40E E points westward. 31 9 2 E = F = mee a = (9:11 101:60kg)(1:80 10 m/s ) = 1:02 10 2 N/C ; e 10 19 C 41E Use Eq. 23-9: ep F = qE = 2 z3 = 2(1:60 10 = 6:6 10 0 15 N : 19 C)(3:6 10 29 C m)(8:99 109 N m2 =C2 ) (25 10 9 m)3 (a) Fe = Ee = (3:0 106 N/C)(1:6 10 (b) Fi = Eqnet = Ee = 4:8 10 13 N: 43E 42E 19 C) = 4:8 10 13 N: Let mg = qE = 2eE . Thus 27 kg)(9 2 E = mg = (6:64 106 10 19:80 m/s ) = 2:03 10 7 N/C : 2e 2(1: C) E should point upward. (a) The magnitude of the force on the particle is given by F = qE , where q is the magnitude of the charge carried by the particle and E is the magnitude of the electric eld at the location of the particle. Thus 6 E = F = 3::0 10 9 N = 1:5 103 N/C : q 2 0 10 C The force points downward and the charge is negative, so the eld points upward. 44E 652 CHAPTER 23 ELECTRIC FIELDS (b) The magnitude of the electrostatic force on a proton is Fe = eE = (1:60 10 19 C)(1:5 103 N/C) = 2:4 10 16 N : A proton is positively charged, so the force is in the same direction as the eld, upward. (c) The magnitude of the gravitational force on the proton is Fg = mg = (1:67 10 The force is downward. (d) The ratio of the forces is 27 kg)(9:8 m/s2 ) = 1:6 10 26 N : Fe = 2:4 10 16 N = 1:5 1010 : Fg 1:64 10 26 N 45E Let mg = jqjE and solve for jqj: 4 jqj = mg = 150:4 N = 2:9 10 2 C : E N/C Since qE must be upward while E is downward, q is negative. 46E (a) (b) (c) 19 C)(1 6 F a = m = eE = (1:60 10:11 10 :40 10 N/C) = 2:46 1017 m/s2 : 31 kg 9 e me v :00 107 m/s t = a = 2346 1017 m/s2 = 1:22 10 : 1 s = 1 at2 = 2 (2:46 1017 m/s2 )(1:22 10 2 10 s : 10 s)2 = 1:83 10 3 m : 47E (a) The magnitude of the force acting on the proton is F = eE , where E is the magnitude of the electric eld. According to Newton's second law the acceleration of the proton is a = F=mp = eE=mp , where mp is the mass of the proton. Thus 19 C)(2 4 a = (1:60 10:67 10 :00 10 N/C) = 1:92 1012 m/s2 : 27 kg 1 CHAPTER 23 ELECTRIC FIELDS 653 (b) Assume the proton starts from rest and use the kinematic equations x = 1 at2 and 2 q p 12 m/s2 )(0:0100 m) = 1:96 105 m/s. v = at to show that v = 2ax = 2(1:92 10 48E 2 (a) Use vf vi2 = vi2 = 2as and a = F=m = eE=m to solve for s: 2 31 6 2 e v2 s = 2vi = meEi = (9:11 10 19kg)(5::00 10 3m/s) = 7:12 10 2 m : a 2 2(1:60 10 C)(1 00 10 N/C) (b) (c) From v2 = 2as :12 10 2 m) s t = v = 2s = 2(700 106 m/s = 2:85 10 8 s : v 5: i K = ( 1 me v2 ) = v2 = 2as = 2eE s 2 1 me v 2 Ki vi2 vi2 me vi2 i 2 19 C)(1: 3 3 = 2(1:60 10 10 3100 10 N/C)(8:00 10 m) = 11:2% : (9:11 kg)(5:00 106 m/s)2 49E (a) 3 6 3 3 3 W = V = D = (1:2 10 m) (1:00 10 kg/m ) = 8:87 10 15 N : 6 6 (b) From W = F = qE = NeE we solve for N , the number of excess electrons: 15 8 W N = eE = (1:60 :87 10 N N/C) = 120 : 10 19 C)(462 When the drop is in equilibrium the force of gravity is balanced by the force of the electric eld: mg = qE , where m is the mass of the drop, q is the charge on the drop, and E is the magnitude of the electric eld. The mass of the drop is given by m = (4=3)r3 , where r is its radius and is its mass density. Thus 3 q = mg = 4rEg E 3 4(1:64 10 6 m)3 (851 kg/m3 )(9:8 m/s2 ) = 8:0 10 = 3(1:92 105 N/C) 50E 19 C 654 CHAPTER 23 ELECTRIC FIELDS and q=e = (8:0 10 51P 19 C)=(1:60 10 19 C) = 5. You need to nd the largest common denominator of the group of numbers listed in the problem. One way to start this is to nd the closest separation between any pair of the numbers. For example, take 13:13 10 19 C 11:50 10 19 C = 1:63 10 19 C. You can easily verify that this value is a common denominator. So this is the value of e that can be deduced. 52P (a) Use s = vt = vt=2: 0 10 2 m) v = 2ts = 2(2::5 10 8 s = 2:7 106 m/s : 1 (b) Use s = 1 at2 and E = F=e = me a=e: 2 2 m)(9:11 31 kg) 0 E = mee a = 2sme = 2(2::60 10 19 C)(1:5 10 8 s)2 = 1:0 103 N/C : et2 (1 10 10 53P (a) F = qE = q(Ex i + Ey j) = (8:00 10 5 C)(3:00 103 N/C) i + (8:00 10 5 C)( 600 N/C) j = (0:240 N) i (0:0480 N) j : p So the magnitude of F is F = (0:240 N)2 + (0:0480 N)2 = 0:245 N, and F makes an angle with the +x direction, where = tan 1 Fy = tan Fx 1 0:0480 N = 11:3 : 0:240 N (b) The coordinates (x; y) at t = 3:00 s are 2 N)(3 s)2 x = 1 ax t2 = Fx t = (0:240 10 :00kg) = 108 m ; 2 2 2m 2(1:0 > 1 a t2 = xay = xFy = (108 m)( 0:0480 N) = 21:6 m : > :y = 2 y ax Fx 0:240 N 8 > > < CHAPTER 23 ELECTRIC FIELDS 655 54P e 10 19 C)(120 F = m = mE = ( 1:60 9:11 10 31 kg N/C) j = 2:11 1013 j (m/s2 ) : e e (b) Since ax = 0, the time t it takes for the x coordinate of the electron to change by 2:0 cm is t = 2:0 cm=(1:5 105 m/s) = 1:3 10 7 s. So vx remains at 1:5 105 m/s, while vy becomes vy = 3:0 103 m/s (2:11 1013 m/s2 )(1:3 10 7 s) = 2:7 106 m/s : Thus v becomes (1:5 105 i 2:7 106 j) m/s. a (a) Take the positive direction to be to the right in the diagram. The acceleration of the proton is ap = eE=mp and the acceleration of the electron is ae = eE=me , where E is the magnitude of the electric eld, mp is the mass of the proton, and me is the mass of the electron. Take the origin to be at the initial position of the proton. Then the coordinate of the proton at time t is x = 1 ap t2 and the coordinate of the electron is x = L + 1 ae t2 . 2 2 They pass each other when their coordinates are the same, or 1 ap t2 = L + 1 ae t2 . This 2 2 means t2 = 2L=(ap ae ) and p x = a ap a L = (eE=meE=meE=m ) L = m mem L p e p) + ( e e+ p 31 kg 11 10 = 9:11 109:31 kg + 1:67 10 27 kg (0:050 m) = 2:7 10 5 m : 55P (a) Suppose the pendulum is at the angle with the vertical. The T force diagram is shown to the right. T is the tension in the thread, mg is the force of gravity, and qE is the force of the electric eld. qE The eld points upward and the charge is positive, so the force is upward. Take the angle shown to be positive. Then the torque on the sphere about the point where the thread is attached to the upper plate is = (mg qE )` sin . If mg > qE then the torque is a restoring torque; it tends to pull the pendulum back to its mg equilibrium position. If the amplitude of the oscillation is small, sin can be replaced by in radians and the torque is = (mg qE )`. The torque is proportional to the angular displacement and the p pendulum moves in simple harmonic motion. Its angular frequency is ! = (mg qE )`=I , where I is the rotational inertia of the pendulum. Since I = m`2 for a simple pendulum, 56P != r (mg qE )` = m`2 r g qE=m ` 656 CHAPTER 23 ELECTRIC FIELDS and the period is ` T = 2! = 2 g qE=m : s If qE > mg the torque is not a restoring torque and the pendulum does not oscillate. (b) The force of the electric eld is now downward and the torque on the pendulum is = (mg + qE )` if the angular displacement is small. The period of oscillation is ` T = 2 g + qE=m : s The electric eld is upward in the diagram and the charge is negative, so the force of the eld on it is downward. The magnitude of the acceleration is a = eE=me , where E is the magnitude of the eld and me is the mass of the electron. Its numerical value is 19 C)(2 3 a = (1:60 10:11 10 :00 10 N/C) = 3:51 1014 m/s2 : 31 kg 9 57P Put the origin of a coordinate system at the initial position of the electron. Take the x axis to be horizontal and positive to the right; take the y axis to be vertical and positive toward the top of the page. The kinematic equations are x = v0 t cos , y = v0 t sin 1 at2 , 2 and vy = v0 sin at. First nd the greatest y coordinate attained by the electron. If it is less than d the electron does not hit the upper plate. If it is greater than d it will hit the upper plate if the corresponding x coordinate is less than L. The greatest y coordinate occurs when vy = 0. This means v0 sin at = 0 or t = (v0 =a) sin and 2 2 2 2 1 2 2 ymax = v0 sin 2 a v0 sin = 1 v0 sin a a2 2 a 6 m/s)2 sin2 45 = (6:00 10 14 2 = 2:56 10 2 m : 2(3:51 10 m/s ) Since this is greater than d (= 2:00 cm) the electron might hit the upper plate. Now nd the x coordinate of the position of the electron when y = d. Since v0 sin = (6:00 106 m/s) sin 45 = 4:24 106 m/s and 2ad = 2(3:51 1014 m/s2 )(0:0200 m) = 1:40 1013 m2 /s2 ; CHAPTER 23 ELECTRIC FIELDS 657 the solution to d = v0 t sin 1 at2 2 is 2 v0 sin2 2ad t= a q 4:24 106 m/s (4:24 106 m/s)2 1:40 1013 m2 /s2 = 6:43 10 9 s : = 14 m/s2 3:51 10 v0 sin q The negative root was used because we want the earliest time for which y = d. The x coordinate is x = v0 t cos = (6:00 106 m/s)(6:43 10 9 s) cos 45 = 2:72 10 2 m : This is less than L so the electron hits the upper plate at x = 2:72 cm. (a) p = (1:5 10 9 C)(6:20 10 6 m) = 9:30 10 15 C m. (b) U = pE ( pE ) = 2pE = 2(9:30 10 15 C m)(1100 N/C) = 2:05 10 59E 58E 11 J: Use = p E. (a) Now = 0 as p k E. (b) Now = pE sin 90 = 2(1:6 10 19 C)(0:78 10 9 m)(3:4 106 N/C) = 8:5 10 (c) Now = 0 again, since p k ( E). 60P 22 N m: W = U = p f E ( p i E) = (pi p f ) E = pE cos 0 pE cos(0 + ) = 2pE cos 0 . The magnitude of the torque acting on the dipole is given by = pE sin , where p is the magnitude of the dipole moment, E is the magnitude of the electric eld, and is the angle between the dipole moment and the eld. It is a restoring torque: it always tends to rotate the dipole moment toward the direction of the electric eld. If is positive the torque is negative and vice-versa. Write = pE sin . If the amplitude of the motion is small we may replace sin with in radians. Thus = pE. Since the magnitude of the torque is proportional to the angle of rotation, the dipole oscillates in simple harmonic motion, just like a torsional pendulum with torsion constant = pE . The angular frequency ! is given by !2 = = pE ; I I 61P 658 CHAPTER 23 ELECTRIC FIELDS where I is the rotational inertia of the dipole. The frequency of oscillation is f = 2! = 21 pE : I 62P r (a) From Eq. 23-38 U= E = [(3:00 i + 4:00 j)(1:24 10 = 1:49 10 26 J : p 30 C m)] [(4000 N/C) i] (b) From Eq. 23-34 = p E = [(3:00 i + 4:00 j)(1:24 10 = ( 1:98 10 26 N m) k : (c) The work done is 30 C m)] [(4000 N/C) i] W = U = ( p E) = (pi pf ) E = [(3:00 i + 4:00 j) ( 4:00 i + 3:00 j)](1:24 10 = 3:47 10 26 J : 63 30 C m)] [(4000 N/C) i] (a) (c) 0.707 (d) 0.21 and 1.9 64 2 : E = 4qd2 (1 + 2 )3=2 0 x=d (a) rst row: 4, 8, 12; second row: 5, 10, 14; thier row: 7, 11, 16; (b) 1:63 10 19 C ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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