Physics exercises_solution: Chapter 21

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Physics exercises_solution: Chapter 21

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 21

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  1. 21.1: mlead  8.00 g and charge  3.20  10 9 C  3.20  10 9 C a) ne  19  2.0  1010.  1.6  10 C 8.00 g n b) nlead  NA   2.33  10 22 and e  8.58  10 13. 207 nlead 21.2: current  20,000 C s and t  100 s  10 4 s Q = It = 2.00 C Q ne  19  1.25  1019. 1.60  10 C 21.3: The mass is primarily protons and neutrons of m  1.67  10 27 kg, so: 70.0 kg np and n   27  4.19  1028 1.67  10 kg About one-half are protons, so n p  2.10  10 28  ne and the charge on the electrons is given by: Q  (1.60  10 19 C)  (2.10  10 28 )  3.35  10 9 C. 21.4: Mass of gold = 17.7 g and the atomic weight of gold is 197 g mol. So the number of atoms N A  mol  (6.02  10 23 )   17.7 g 197 g mol   5.41  10 22 . a) np  79  5.41  1022  4.27  1024 q  np  1.60  1019 C  6.83  105 C b) ne  n p  4.27  10 24. 21.5: 1.80 mol  1.80  6.02  1023 H atoms  1.08  1024 electrons. charge  1.08  1024  1.60  1019 C  1.73  105 C. 21.6: First find the total charge on the spheres: 1 q2 F 2  q  4πε0 Fr 2  4πε0 (4.57  10 21 )(0.2) 2  1.43  1016 C 4πε0 r And therefore, the total number of electrons required is n  q e  1.43  1016 C 1.60  1019 C  890.
  2. 21.7: a) Using Coulomb’s Law for equal charges, we find: 1 q2 F  0.220 N   q  5.5  1013 C 2  7.42  10 7 C. 4πε0 (0.150 m)2 b) When one charge is four times the other, we have: 1 4q 2 F  0.220 N   q  1.375  1013 C 2  3.71  10 7 C 4πε0 (0.150 m)2 So one charge is 3.71  10 7 C, and the other is 1.484  10 6 C. 21.8: a) The total number of electrons on each sphere equals the number of protons. 0.0250 kg ne  np  13  N A   7.25  1024. 0.026982 kg mol b) For a force of 1.00  10 4 N to act between the spheres, 1 q2 F  104 N  2  q  4πε0 (104 N) (0.08 m)2  8.43  10 4 C. 4πε0 r   ne  q e  5.27  1015  c) ne is 7.27  10 10 of the total number. 21.9: The force of gravity must equal the electric force. 1 q2 1 (1.60  1019 C) 2 mg   r2   25.8 m 2  r  5.08 m. 4πε0 r 2 4πε0 (9.11  1031 kg)(9.8 m s) 21.10: a) Rubbing the glass rod removes electrons from it, since it becomes positive. 7.50 nC  (7.50  10 9 C) (6.25  1018 electrons C)  4.69  1010 electrons (4.69  1010 electrons) (9.11  10 31 kg electron)  4.27  10  20 kg. The rods mass decreases by 4.27  10 20 kg. b) The number of electrons transferred is the same, but they are added to the mass of the plastic rod, which increases by 4.27  10 20 kg.
  3.   21.11: F2 is in the  x - direction, so F1 must be in the  x - direction and q1 is positive. qq q 2 q3 F1  F2 , k 12 3  k 2 r13 r23 q1  0.0200 0.0400 q 2  0.750 nC 2 1 q1q2 1 (0.550  106 C)q2 21.12: a) F  0.200 N  4πε0 r 2 4πε0 (0.30 m)2  q 2   3.64  10 6 C. b) F  0.200 N, and is attractive. 21.13: Since the charges are equal in sign the force is repulsive and of magnitude: kq 2 (3.50  106 C) 2 F 2   0.172 N r 4πε0 (0.800 m)2 21.14: We only need the y-components, and each charge contributes equally. 1 (2.0  106 C) (4  106 C) F sin α  0.173 N (since sin α  0.6). 4πε0 (0.500 m)2 Therefore, the total force is 2 F  0.35 N , downward.   21.15: F2 and F3 are both in the  x-direction. q1q2 qq F2  k 2  6.749  10 5 N, F3  k 1 2 3  1.124  10 4 N r12 r13 F  F2  F3  1.8  104 N, in the  x-direction.
  4. (9  10 9 N  m 2 C 2 ) (20.  10 6 C) (2.0  10 6 C) 21.16: F21   0.100 N 0.60m 2 FQ1 is equal and opposite to F1Q (Ex. 21.4), so F  Q1 x  0.23 N F  Q1 y  0.17 N Overall: Fx  0.23 N Fy  0.100 N  0.17 N  0.27 N The magnitude of the total force is 0.23 N   0.27 N   0.35 N. The direction of 2 2 the force, as measured from the +y axis is 0.23 θ tan 1  40 0.27  21.17: F2 is in the  x  direction. qq F2  k 1 2 2  3.37 N, so F2 x  3.37 N r12 Fx  F2 x  F3 x and Fx  7.00 N F3 x  Fx  F2 x  7.00 N  3.37 N  10.37 N For F3 x to be negative, q3 must be on the –x-axis. q1q3 k q1q3 F3  k 2 , so x   0.144 m, so x  0.144 m x F3
  5. 21.18: The charge q3 must be to the right of the origin; otherwise both q 2 and q3 would exert forces in the + x direction. Calculating the magnitude of the two forces: 1 q1q2 (9  109 N  m 2 C 2 )(3.00  10 6 C)(5.00  10 6 C) F21  2  4πε0 r12 (0.200 m) 2  3.375 N in the  x direction. (9  10 9 N  m 3 C 2 ) (3.00  10 6 C) (8.00  10 6 C) F31  2 r13 0.216 N  m 2  2 in the  x direction r13 We need F21  F31  7.00 N : 0.216 N  m 2 3.375 N  2  7.00 N r13 0.216 N  m 2 r  2 13  0.0208 m 2 3.375 N  7.00 N r13  0.144 m to the right of the origin    21.19: F  F1  F2 and F  F2  F1 since they are acting in the same direction at y   0.400 m so, 1  1.50  10 9 C 3.20  10 9 C  F (5.00  10  9 C)  6  (0.200 m) 2  (0.400 m) 2   2.59  10 N downward.  4πε0      21.20: F  F1  F2 and F  F1  F2 since they are acting in opposite directions at x = 0 so, 1 9  4.00  10 9 C 5.00  10 9 C  6 F (6.00  10 C)   (0.200 m) 2  (0.300 m)2   2.4  10 N to the right.  4πε0  
  6. 21.21: a) 1 qQ 1 2qQa b) Fx  0, Fy  2 sin θ 4πε 0 (a  x ) 2 2 4πε 0 (a  x 2 ) 3 2 2 1 2qQ c) At x  0, Fy  in the  y direction. 4πε0 a 2 d)
  7. 21.22: a) 1 qQ 1  2qQx b) Fx  2 cos θ  , Fy  0 4πε 0 (a  x ) 2 2 4πε 0 (a  x 2 ) 3 / 2 2 c) At x = 0, F = 0. d) 21.23: b) F  1 q2 4πε0 2 L2  2 1 q2 4πε0 L2   1 2 2 1 q2  4πε0 2 L2 at an angle of 45 below the positive x-axis 1 q 1 (3.00  109 C) 21.24: a) E   432 N C , down toward the particle. 4πε0 r 2 4πε0 (0.250 m)2 1 q 1 (3.00  109 C) b) E  12.00 N C   r  1.50 m. 4πε0 r 2 4πε0 (12.0 N C)
  8. 21.25: Let +x-direction be to the right. Find a x : v0 x  1.50  103 m s , vx  1.50  103 m s , t  2.65  106 s, ax  ? vx  v0 x  axt gives ax  1.132  109 m s 2 Fx  max  7.516  1018 N   F is to the left  x - direction , charge is positive, so E is to the left. E  F q  (7.516  1018 N) 2 (1.602  10 19  C)  23.5 N C 21.26: (a) x  1 at 2 2 2x 2(4.50 m) 2 a 2   1.00  1012 m s t (3.00  10 s) -6 2 2 F ma (9.11  10 31 kg) (1.00  1012 m s ) E   q q 1.6  10 19 C  5.69 N C The force is up, so the electric field must be downward since the electron is negative. (b) The electron’s acceleration is ~ 1011 g, so gravity must be negligibly small compared to the electrical force. (0.00145 kg) (9.8 m s 2 ) 21.27: a) q E  mg  q   2.19  10 5 C, sign is negative. 650 N C (1.67  1027 kg) (9.8 m s 2 ) b) qE  mg  E  19  1.02  10 7 N / C, upward. 1.60  10 C 1 q 1 (26  1.60  1019 C) 21.28: a) E  10  1.04  1011 N C . 4πε0 r 2 4πε0 (6.00  10 m) 2 1 q 1 (1.60  1019 C) b) Eproton   11  5.15  1011 N C . 4πε0 r 2 4πε0 (5.29  10 m) 2 21.29: a) q  55.0  10 6 C, and F is downward with magnitude 6.20  10 9 N. Therefore, E  F q  1.13  10 4 N C, upward. b) If a copper nucleus is placed at that point, it feels an upward force of magnitude F  qE  29  1.6  10 19 C  1.13  10 4 N C  5.24  10 22 N.
  9. 21.30: a) The electric field of the Earth points toward the ground, so a NEGATIVE charge will hover above the surface. 2 (60.0 kg) (9.8 m s ) mg  qE  q    3.92 C. 150 N C 1 q2 1 (3.92 C) 2 b) F    1.38  107 N. The magnitude of the charge is 4πε0 r 2 4πε0 (100.00 m)2 too great for practical use. 21.31: a) Passing between the charged plates the electron feels a force upward, and just misses the top plate. The distance it travels in the y-direction is 0.005 m. Time of flight  t  1.60 . 106 m s  1.25  10 8 s and initial y-velocity is zero. Now, 0 0200 m 2 y  v0 yt  1 at 2 so 0.005 m  1 a(1.25  108 s) 2  a  6.40  1013 m s . But also 2 2 ( 9.11  10 31 kg)( 6.40  1013 m s 2 ) a F m  eE me E 1.60  10 19 C  364 N C . b) Since the proton is more massive, it will accelerate less, and NOT hit the plates. To find the vertical displacement when it exits the plates, we use the kinematic equations again: 1 1 eE y  at 2  (1.25  10 8 s) 2  2.73  10 6 m. 2 2 mp c) As mention in b), the proton will not hit one of the plates because although the electric force felt by the proton is the same as the electron felt, a smaller acceleration results for the more massive proton. d) The acceleration produced by the electric force is much greater than g; it is reasonable to ignore gravity.
  10. 21.32: a)  q1 ˆ (9  10 9 N  m 2 C 2 ) (5.00  10 9 C) E1  j  (2.813  10 4 N C) ˆ j 4πε 0 r1 2 0.0400 m  2  q 2 (9  10 9 N  m 2 C 2 ) (3.00  10 9 C) E2  2   1.08  10 4 N C r2 0.0300m   (0.0400 m) 2 2  The angle of E 2 , measured from the x - axis, is 180  tan 1  4.00 cm 3.00 cm   126.9 Thus  ˆ E 2  (1.080  104 N C) ( i cos 126.9  ˆ sin 126.9) j ˆ  (  6.485  103 N C) i  (8.64  103 N C) ˆ j b) The resultant field is   ˆ E1  E 2  (  6.485  103 N C) i  (  2.813  104 N / C  8.64  103 N C) ˆ j ˆ  (  6.485  103 N / C) i  (1.95  104 N C) ˆj 21.33: Let  x be to the right and  y be downward. Use the horizontal motion to find the time when the electron emerges from the field: x  x0  0.0200 m, a x  0, v 0 x  1.60  10 6 m s , t  ? x  x0  v0 x t  1 a x t 2 gives t  1.25  10 8 s 2 v x  1.60  10 6 m s y  y 0  0.0050 m, v0y  0, t  1.25  10 8 s, v y  ?  v0 y  v y  y  y0     t gives v y  8.00  10 5 m s   2  v  v x  v y  1.79  10 6 m s 2 2  ˆ 21.34: a) E  11 N Ci  14 N Cˆ, so E  (11)2  (14) 2  17.8 N C. j θ  tan 1 (  14 11)   51.8, so θ  128 counterclockwise from the x-axis   b) F  E q so F  (17.8 N C) (2.5  10 9 C)  4.45  108 N, i) at  52 (repulsive) ii) at  128 (repulsive).
  11. 2 21.35: a) Fg  me g  (9.11  10 31 kg) (9.8 m s )  8.93  10 30 N. Fe  eE  (1.60  1019 C) (1.00  104 N C)  1.60  1015 N. Yes, ok to neglect Fg because Fe  Fg . b) E  10 4 N C  Fe  1.6  10 15 N  mg  m  1.63  10 16 kg  m  1.79  1014 me . c) No. The field is uniform. 1 2 1 eE 2 2(0.0160 m) (1.67  10 27 kg ) 21.36: a) x  at  t E  148 N C . 2 2 mp (1.60  10 19 C) (1.50  10 6 s ) 2 eE b) v  v0  at  t  2.13  10 4 m s . mp   1.35  π  1  12  π 2ˆ 2 ˆ 21.37: a) tan 1     ,  r  ˆ b) tan    , r  j ˆ i j  0  2  .2  4 2 2  2.6  ˆ ˆ c) tan 1    1.10   1.97 radians  112.9, r   0.39i  0.92 j (Second quadrant).  ˆ   21.38: a) E  614 N C , F  qE  9.82  10 17 N. b) F  e 2 4πε0 (1.0  1010 ) 2  2.3  10 8 N. c) Part (b) >> Part (a), so the electron hardly notices the electric field. A person in the electric field should notice nothing if physiological effects are based solely on magnitude.
  12. 21.39: a) Let  x be east.   E is west and q is negative, so F is east and the electron speeds up. Fx | q | E  (1.602  1019 C) (1.50 V m)  2.403  1019 N ax  Fx m  (2.403  1019 N) (9.109  10 31kg)   2.638  1011 m s 2 v0 x   4.50  105 m s , ax   2.638  1011 m s 2 , x  x0  0.375 m, vx  ? vx  v0 x  2ax ( x  x0 ) gives vx  6.33  105 m s 2 2  b) q  0 so F is west and the proton slows down. Fx   | q | E   (1.602  1019 C) (1.50 V m)   2.403  1019 N 2 ax  Fx m  (  2.403  1019 N) (1.673  10 27 kg )   1.436  108 m s 2 v0 x   1.90  104 m s , a x   1.436  108 m s , x  x0  0.375 m, vx  ? v 2 x  v 2 0 x  2a x ( x  x0 ) gives vx  1.59  10 4 m s 21.40: Point charges q1 (0.500 nC) and q 2 (8.00 nC) are separated by x  1.20 m. The kq1 kq 2 electric field is zero when E1  E2  r12  (1.20  r1 ) 2  q2 r12  q1 (1.2  r1 ) 2  q1r12  2q1 (1.2)r1  1.22 q1  (q2  q1 )r12  2(1.2)q1r1  (1.2) 2 q1  0 or 7.5r12  1.2r1  0.72  r1   0.24,  0.4 r1  0.24 is the point between.
  13. 21.41: Two positive charges, q , are on the x-axis a distance a from the origin. a) Halfway between them, E  0.  1  q q     ( a  x) 2  ( a  x) 2  , | x |  a   4πε0    1   q q  b) At any position x, E     ( a  x) 2  ( a  x ) 2  , x  a   4πε0    1  q q     ( a  x) 2  ( a  x ) 2  , x   a   4πε0    For graph, see below. 21.42: The point where the two fields cancel each other will have to be closer to the negative charge, because it is smaller. Also, it cant’t be between the two, since the two fields would then act in the same direction. We could use Coulomb’s law to calculate the actual values, but a simpler way is to note that the 8.00 nC charge is twice as large as the –4.00 nC Charge. The zero point will therefore have to be a factor of 2 farther from the 8.00 nC charge for the two fields to have equal magnitude. Calling x the distance from the –4.00 nC charge: 1.20  x  2 x x  2.90 m
  14. 21.43: a) Point charge q1 (2.00 nC) is at the origin and q 2 (  5.00 nC ) is at x  0.800 m. k | q1 | k | q2 | i) At x  0.200 m, E  2   575 N C right. (0.200 m) (0.600 m) 2 k | q2 | k | q1 | ii) At x  1.20 m, E  2   269 N C left. (0.400 m) (1.20 m) 2 k | q1 | k | q2 | iii) At x   0.200 m, E  2   405 N C left. (0.200 m) (1.00 m) 2 b) F   eE i) F  1.6  1019 C  575 N C  9.2  1017 N left, ii) F  1.6  1019 C  269 N C  4.3  1017 N right, iii) F  1.6  1019  405  6.48  1017 N right. 21.44: A positive and negative charge, of equal magnitude q , are on the x-axis, a distance a from the origin. 1 2q a) Halfway between them, E  , to the left. 4πε0 a 2  1  q q     (a  x) 2  (a  x) 2  , | x |  a   4πε0    1   q q  b) At any position x, E     (a  x) 2  (a  x) 2  , x  a   4πε0    1  q q     (a  x) 2  (a  x) 2  , x   a   4πε0    with “+” to the right. This is graphed below.
  15. 21.45: a) At the origin, E  0. b) At x  0.3 m, y  0 :  1  1 1 ˆ (6.00  10  9 C)  ˆ E 4πε0  (0.15 m) 2  (0.45 m) 2 i  2667 i N C .    c) At x  0.15 m, y  0.4 m :  1  1 ˆ 1 0.3 ˆ 1 0.4 ˆ  E (6.00  10  9 C)   (0.4 m) j i j 4πε0  2 2 (0.5 m) 0.5 (0.5 m) 0.5  2    E  (129.6iˆ  510.3 ˆ) N C  E  526.5 N C and θ  75.7 down from the x-axis. j  0.2   2(6.00  10 9 C)    1  0.25   1382 ˆ N C d) x  0, y  0.2 m : E  j 4πε0 (0.25 m) 2 21.46: Calculate in vector form the electric field for each charge, and add them.   1 (6.00  109 C) ˆ E  ˆ i  150i N C 4πε0 (0.6 m) 2   1 (4.00  10 9 C) 1 ˆ 1 ˆ ˆ ˆ E  4πε0  (1.00 m) 2 (0.6)i  (1.00 m) 2 (0.8) j   21.6i  28.8 j N C     28.8   E  (128.4) 2  (28.8) 2  131.6 N C , at θ  tan 1     12.6 up from  128.4   x axis.  1 2(6.0  109 C) ˆ 21.47: a) At the origin, E   ˆ i  4800i N C . 4πε0 (0.15 m) 2 b) At x  0.3 m, y  0 :  1  1 1 ˆ (6.0  10  9 C)  ˆ E 4πε0  (0.15 m) 2  (0.45 m) 2  i  2133i N C .    c) At x  0.15 m, y  0.4 m :  1  1 ˆ 1 0.3 ˆ 1 0.4 ˆ  E (6.0  10 9 C)  (0.4 m) j i j 4πε0  2 2 (0.5 m) 0.5 (0.5 m) 0.5  2    E  (129.6i  164.5 ˆ) N C  E  209 N C and θ  232 clockwise from ˆ j  x - axis.  1 2(6.00  109 C)  0.15  ˆ d) x  0, y  0.2 m : E y  0, E   0.25   1037i N C 4πε0 (0.25 m)2
  16. λ 1.5  1010 C m 21.48: For a long straight wire, E  r  1.08 m. 2πε0 r 2πε0 (2.5 N C) 21.49: a) For a wire of length 2a centered at the origin and lying along the y-axis, the electric field is given by Eq. (21.10).  1 λ E ˆ i 2πε0 x x 2 a 2  1 b) For an infinite line of charge:  λ ˆ E i 2πε0 x Graphs of electric field versus position for both are shown below. 21.50: For a ring of charge, the electric field is given by Eq. (21.8).  1 Qx a) E  ˆ i so with 4πε0 ( x  a 2 )3 2 2  ˆ Q  0.125  10 9 C, a  0.025 m and x  0.4 m  E  7.0i N C .    ˆ ˆ b) Fon ring   Fon q   q E   (  2.50  106 C) (7.0i N C)  1.75  10-5 i N.
  17. 21.51: For a uniformly charged disk, the electric field is given by Eq. (21.11):  σ  1  E 1  iˆ 2 ε0  R x 1 2 2   The x -component of the electric field is shown below. 21.52: The earth’s electric field is 150 N C , directly downward. So, σ 2 E  150   σ  300ε0  2.66  10  9 C m , and is negative. 2 ε0 σ 21.53: For an infinite plane sheet, E is constant and is given by E  directed 2ε0 perpendicular to the surface. 2 e  C   100 cm  9 C σ  2.5  10 6 2   1.6  1019      1 m    4  10 m 2  cm  e    4  10 9 C so E  m2  226 N C directed toward the surface. 2 ε0 21.54: By superposition we can add the electric fields from two parallel sheets of charge. a) E  0. b) E  0. σ σ c) E  2  , directed downward. 2 ε0 ε0
  18. 21.55: 21.56: The field appears like that of a point charge a long way from the disk and an infinite plane close to the disk’s center. The field is symmetrical on the right and left (not shown).
  19. 21.57: An infinite line of charge has a radial field in the plane through the wire, and constant in the plane of the wire, mirror-imaged about the wire: Cross section through the wire: Plane of the wire: Length of vector does not depend on angle. Length of vector gets shorter at points further away from wire. 21.58: a) Since field lines pass from positive charges and toward negative charges, we can deduce that the top charge is positive, middle is negative, and bottom is positive. b) The electric field is the smallest on the horizontal line through the middle charge, at two positions on either side where the field lines are least dense. Here the y- components of the field are cancelled between the positive charges and the negative charge cancels the x-component of the field from the two positive charges. 21.59: a) p  qd  (4.5  109 C)(0.0031 m)  1.4  1011 C  m , in the direction from and towards q 2 .  b) If E is at 36.9, and the torque τ  pE sin  , then: τ 7.2  109 N  m E   856.5 N C . p sin  (1.4  1011 C  m) sin 36.9
  20. 21.60: a) d  p q  (8.9  1030 C  m) (1.6  1019 C)  5.56  1011 m. b) τ max  pE  (8.9  10 30 C  m)(6.0  105 N C)  5.34  1024 N  m. Maximum torque: 21.61: a) Changing the orientation of a dipole from parallel to perpendicular yields: U  U f  U i   ( pE cos 90  pE cos 0)   (5.0  1030 C  m)(1.6  10 6 N C)   8  1024 J. 3 2(8  1024 J) b) kT  8  10 24 J  T   0.384 K. 2 3(1.38  10 23 J K ) p 9 6.17  1030 C  m 21.62: Edipole ( x)   Edipole (3.00  10 m)   4.11 2πε0 x 3 20 (3.0  10 9 m)3  106 N C . The electric force F  qE  (1.60  1019 C)(4.11  106 N C)  6.58  1013 N and is toward the water molecule (negative x-direction). 1 1 ( y  d 2) 2  ( y  d 2) 2 2 yd 21.63: a)    2 ( y  d 2) 2 ( y  d 2) 2 ( y  d 4) 2 2 2 ( y  d 2 4) 2 q 2 yd qd y p  Ey    4πε0 ( y  d 4 ) 2 2 2 2πε0 ( y  d 4) 2 2 2 2πε0 y 3 b) This also gives the correct expression for E y since y appears in the full expression’s denominator squared, so the signs carry through correctly.
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