21.1: mlead  8.00 g and charge  3.20  10 9 C a) ne  b) nlead 21.2:  3.20  10 9 C  2.0  1010 19  1.6  10 C n 8.00 g  NA   2.33  10 22 and e  8.58  10 13 207 nlead current  20,000 C s and t  100 s  10 4 s Q = It = 2.00 C Q ne   1.25  1019 19 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   4.19  1028  27 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 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: q2 F  q  4πε0 Fr  4πε0 (4.57  10 21 )(0.2)  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 21.7: a) Using Coulomb’s Law for equal charges, we find: q2 F  0.220 N   q  5.5  1013 C  7.42  10 C 4πε0 (0.150 m)2 b) When one charge is four times the other, we have: 4q F  0.220 N   q  1.375  1013 C  3.71  10 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 ne  np  13  N A  0.0250 kg  7.25  1024 0.026982 kg mol b) For a force of 1.00  10 N to act between the spheres, F  104 N  q2  q  4πε0 (104 N) (0.08 m)2  8.43  10 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 q2 (1.60  1019 C) 2   mg  r  25.8 m  r  5.08 m 4πε0 r 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   21.11: F2 is in the  x - direction, so F1 must be in the  x - direction and q1 is positive q q3 qq F1  F2 , k 12  k r13 r23 q1  0.0200 0.0400 q  0.750 nC 21.12: a) F q1q2 (0.550  106 C)q2 200 N   4πε0 r 4πε0 (0.30 m)2  q   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 (3.50  106 C) F   0.172 N 4πε0 (0.800 m)2 r 21.14: We only need the y-components, and each charge contributes equally (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 F  0.35 N , downward   21.15: F2 and F3 are both in the  x-direction F2  k q1q2 qq  6.749  10 N, F3  k  1.124  10 N r12 r13 F  F2  F3  1.8  104 N, in the  x-direction 21.16: F21  (9  10 N  m C ) (20  10 6 C) (2.0  10 6 C) 0.60m 2  0.100 N FQ1 is equal and opposite to F1Q (Ex 21.4), so F  F  Q1 x Q1 y  0.23 N  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 the force, as measured from the +y axis is 0.23 θ tan 1  40 0.27 2  21.17: F2 is in the  x  direction qq F2  k 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 F3  k q1q3 k q1q3 , so x   0.144 m, so x  0.144 m x F3 21.18: The charge q3 must be to the right of the origin; otherwise both q and q3 would exert forces in the + x direction Calculating the magnitude of the two forces: q1q2 (9  109 N  m C )(3.00  10 6 C)(5.00  10 6 C)  F21  4πε0 r122 (0.200 m)  3.375 N in the  x direction (9  10 N  m C ) (3.00  10 6 C) (8.00  10 6 C) F31  r132 0.216 N  m in the  x direction r132 We need F21  F31  7.00 N :  0.216 N  m 3.375 N   7.00 N r132 0.216 N  m r   0.0208 m 3.375 N  7.00 N 13 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, F  1.50  10 9 C 3.20  10 9 C    2.59  10  N downward  (5.00  10  C)  2  4πε0 ( 200 m) ( 400 m)      21.20: F  F1  F2 and F  F1  F2 since they are acting in opposite directions at x = so,  4.00  10 9 C 5.00  10 9 C  9   2.4  10  N to the right  (6.00  10 C)  F 2  4πε0 (0.300 m)   (0.200 m) 21.21: a) 1 qQ 2qQa sin θ 2 4πε (a  x ) 4πε (a  x ) 2qQ c) At x  0, Fy  in the  y direction 4πε0 a b) Fx  0, Fy  d) 21.22: a) b) Fx  2 1 qQ  2qQx cos θ  , Fy  2 4πε (a  x ) 4πε (a  x ) / c) At x = 0, F = d) 21.23:   q2 q2 q2 2    at an angle of 45 below the 4πε0 L2 4πε0 L2 4πε0 L2 positive x-axis b) F  21.24: a) E  q (3.00  109 C)  432 N C , down toward the particle 4πε0 r 4πε0 (0.250 m)2 q (3.00  109 C)  r b) E  12.00 N C   1.50 m 4πε0 r 4πε0 (12.0 N C) 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 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  12 at 2(4.50 m) 2x a   1.00  1012 m s -6 t (3.00  10 s) E F ma (9.11  10 31 kg) (1.00  1012 m s )   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 ) 21.27: a) q E  mg  q   2.19  10 C, sign is negative 650 N C (1.67  1027 kg) (9.8 m s ) b) qE  mg  E   1.02  10 N / C, upward 19 1.60  10 C 21.28: a) b) q (26  1.60  1019 C)  E  1.04  1011 N C 2 10 4πε0 r 4πε0 (6.00  10 m) Eproton q (1.60  1019 C)    5.15  1011 N C 2 11 4πε0 r 4πε0 (5.29  10 m) 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 21.30: a) The electric field of the Earth points toward the ground, so a NEGATIVE charge will hover above the surface (60.0 kg) (9.8 m s ) mg  qE  q    3.92 C 150 N C q2 (3.92 C)   1.38  107 N The magnitude of the charge is 4πε0 r 4πε0 (100.00 m)2 too great for practical use b) F  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 m  t  1.600.0200  1.25  10 8 s and initial y-velocity is zero Now, 10 m s y  v0 yt  12 at so 0.005 m  12 a(1.25  108 s)  a  6.40  1013 m s But also a F m  eE me E ( 9.11  10 31 kg)( 6.40  1013 m s ) 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 eE y  at  (1.25  10 8 s)  2.73  10 6 m 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 21.32: a)  q1 ˆ (9  10 N  m C ) (5.00  10 9 C) E1  j  (2.813  10 N C) ˆj 2 0.0400 m  4πε r1  q (9  10 N  m C ) (3.00  10 9 C) E  22   1.08  10 N C 2 r2 0.0300m   (0.0400 m)  The angle of E , measured from the x - axis, is 180  tan 1  4.00 cm 3.00 cm   126.9 Thus  E  (1.080  104 N C) ( iˆ cos 126.9  ˆj sin 126.9)  (  6.485  103 N C) iˆ  (8.64  103 N C) ˆj b) The resultant field is   E1  E  (  6.485  103 N C) iˆ  (  2.813  104 N / C  8.64  103 N C) ˆj  (  6.485  103 N / C) iˆ  (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 x  1.60  10 m s , t  ? x  x0  v0 x t  12 a x t gives t  1.25  10 8 s v x  1.60  10 m s y  y  0.0050 m, v0y  0, t  1.25  10 8 s, v y  ?  v0 y  v y y  y      t gives v y  8.00  10 m s  v  v x2  v y2  1.79  10 m s  21.34: a) E  11 N Ciˆ  14 N Cˆj , so E  (11)2  (14)  17.8 N C 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) ... downward   21. 15: F2 and F3 are both in the  x-direction F2  k q1q2 qq  6.749  10 N, F3  k  1.124  10 N r12 r13 F  F2  F3  1.8  104 N, in the  x-direction 21. 16: F21  (9  10... 10 6 C)  F21  4πε0 r122 (0.200 m)  3.375 N in the  x direction (9  10 N  m C ) (3.00  10 6 C) (8.00  10 6 C) F31  r132 0 .216 N  m in the  x direction r132 We need F21  F31  7.00... need F21  F31  7.00 N :  0 .216 N  m 3.375 N   7.00 N r132 0 .216 N  m r   0.0208 m 3.375 N  7.00 N 13 r13  0.144 m to the right of the origin    21. 19: F  F1  F2 and F  F2 