Physics 104 21.8. Assignment 21 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: NA 602 1023 atoms/mol. q nee, where ne 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 kg 24 ne np (13)( N A ) 10 electrons. 0.026982 kg/mol 1 q2 . This gives 4e0 r 2 (b) For a force of 100 104 N to act between the spheres, F 100 104 N q 4e0 (100 104 N)(0800 m)2 843 104 C. The number of electrons removed from one sphere and added to the other is ne q /e 527 1015 electrons 21.9. (c) ne/ne 727 1010. EVALUATE: When ordinary objects receive a net charge the fractional change in the total number of electrons in the object is very small. IDENTIFY: Apply Coulomb’s law. SET UP: Consider the force on one of the spheres. (a) EXECUTE: q1 q2 q F 1 q1q2 q2 F 0220 N so q r 0150 m 742 107 C (on each) 2 9 2 2 4 e0 r (1/4 e0 ) 4 e0r 2 8988 10 N m /C (b) q2 4q1 F 1 q1q2 4q12 F F 1r 1 (742 107 C) 371 107 C so q1 r 4 e0 r 2 4(1/4 e0 ) 2 (1/4 e0 ) 2 4 e0r 2 And then q2 4q1 148 106 C EVALUATE: The force on one sphere is the same magnitude as the force on the other sphere, whether the spheres have equal charges or not. 21.15.IDENTIFY: Apply Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite directions. SET UP: Like charges repel and unlike charges attract. q q EXECUTE: The force F2 that q2 exerts on q3 has magnitude F2 k 2 2 3 and is in the x-direction. r2 F1 must be in the x-direction, so q1 must be positive. F1 F2 gives k q1 q3 r12 k q2 q3 r22 . 2 2 r 200 cm q1 q2 1 300 nC 0750 nC. 400 cm r2 EVALUATE: The result for the magnitude of q1 doesn’t depend on the magnitude of q2 . 21.17.IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on q1. SET UP: Like charges repel and unlike charges attract, so F2 and F3 are both in the x-direction. EXECUTE: F2 k q1q2 r122 6749 102 5 N, F3 k q1q3 r132 F 18 104 N and is in the x-direction. 1 1124 104 N. F F2 F3 18 104 N. EVALUATE: Comparing our results to those in Example 21.3, we see that F1 on 3 F3 on 1, as required by Newton’s third law. 21.21.IDENTIFY: Apply Coulomb’s law to calculate the force each of the two charges exerts on the third charge. Add these forces as vectors. SET UP: The three charges are placed as shown in Figure 21.21a. Figure 21.21a EXECUTE: Like charges repel and unlike attract, so the free-body diagram for q3 is as shown in Figure 21.21b. 1 q1q3 F1 4 e0 r132 F2 1 q2q3 2 4 e0 r23 Figure 21.21b F1 8988 109 N m 2 /C2 F2 8988 109 N m 2 /C2 (150 109 C)(500 109 C) (0200 m) 2 (320 109 C)(500 109 C) (0400 m) 2 1685 106 N 8988 107 N The resultant force is R F1 F2 Rx 0 Ry F1 F2 1685 106 N 8988 107 N 258 106 N The resultant force has magnitude 258 106 N and is in the y-direction. EVALUATE: The force between q1 and q3 is attractive and the force between q2 and q3 is replusive. 21.25.IDENTIFY: F q E. Since the field is uniform, the force and acceleration are constant and we can use a constant acceleration equation to find the final speed. SET UP: A proton has charge e and mass 167 1027 kg. EXECUTE: (a) F 160 1019 C275 103 N/C 440 1016 N (b) a F 440 1016 N 263 1011 m/s 2 m 167 1027 kg (c) vx v0 x axt gives v (263 1011 m/s2 )(100 106 s) 263 105 m/s EVALUATE: The acceleration is very large and the gravity force on the proton can be ignored. 21.29.(a) IDENTIFY: Eq. (21.4) relates the electric field, charge of the particle, and the force on the particle. If the particle is to remain stationary the net force on it must be zero. SET UP: The free-body diagram for the particle is sketched in Figure 21.29. The weight is mg, downward. For the net force to be zero the force exerted by the electric field must be upward. The electric field is downward. Since the electric field and the electric force are in opposite directions the charge of the particle is negative. 2 mg q E Figure 21.29 mg (145 102 3 kg)(980 m/s 2 ) 219 105 C and q 219C E 650 N/C (b) SET UP: The electrical force has magnitude FE q E eE The weight of a proton is w mg FE w so eE mg q EXECUTE: mg (1673 1027 kg)(980 m/s 2 ) 102 107 N/C e 1602 1019 C This is a very small electric field. EVALUATE: In both cases q E mg and E (m/ q ) g In part (b) the m/ q ratio is much smaller ( 108 ) EXECUTE: E than in part (a) ( 102 ) so E is much smaller in (b). For subatomic particles gravity can usually be ignored compared to electric forces. q 21.31.IDENTIFY: For a point charge, E k 2 The net field is the vector sum of the fields produced by each charge. A r charge q in an electric field E experiences a force F qE SET UP: The electric field of a negative charge is directed toward the charge. Point A is 0.100 m from q2 and 0.150 m from q1. Point B is 0.100 m from q1 and 0.350 m from q2 . EXECUTE: (a) The electric fields at point A due to the charges are shown in Figure 21.31a. q 625 109 C E1 k 21 899 109 N m 2 /C 2 250 103 N/C rA1 (0150 m) 2 E2 k q2 rA22 (899 109 N m 2 /C2 ) 125 109 C (0100 m) 2 1124 104 N/C Since the two fields are in opposite directions, we subtract their magnitudes to find the net field. E E2 E1 874 103 N/C, to the right. (b) The electric fields at point B are shown in Figure 21.31b. q 625 109 C E1 k 21 899 109 N m 2 /C2 5619 103 N/C 2 rB1 (0100 m) E2 k q2 rB22 (899 109 N m 2 /C 2 ) 125 109 C (0350 m) 2 917 102 N/C Since the fields are in the same direction, we add their magnitudes to find the net field. E E1 E2 654 103 N/C, to the right. (c) At A, E 874 103 N/C, to the right. The force on a proton placed at this point would be F qE (160 1019 C)(874 103 N/C) 140 1015 N, to the right. EVALUATE: A proton has positive charge so the force that an electric field exerts on it is in the same direction as the field. Figure 21.31 3 21.32.IDENTIFY: The electric force is F qE . SET UP: The gravity force (weight) has magnitude w mg and is downward. EXECUTE: (a) To balance the weight the electric force must be upward. The electric field is downward, so for an upward force the charge q of the person must be negative. w F gives mg q E and q mg (60 kg)(980 m/s 2 ) 39 C. E 150 N/C (b) F k 21.33. qq 899 109 N m 2 /C 2 (39 C) 2 14 107 N. The repulsive force is immense and this is not a r2 (100 m) 2 feasible means of flight. EVALUATE: The net charge of charged objects is typically much less than 1 C. 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. (a) SET UP: The motion is sketched in Figure 21.33a. For an electron q e Figure 21.33a F qE and q negative gives that F and E are in opposite directions, so F is upward. The free-body diagram for the electron is given in Figure 21.33b. EXECUTE: Fy ma y eE ma Figure 21.33b Solve the kinematics to find the acceleration of the electron: Just misses upper plate says that x x0 200 cm when y y0 0500 cm x-component v0 x v0 160 106 m/s, ax 0,x x0 00200 m, t ? x x0 v0 xt 12 a xt 2 x x0 00200 m 125 108 s v0 x 160 106 m/s In this same time t the electron travels 0.0050 m vertically: y-component t t 125 108 s, v0 y 0, y y0 00050 m, a y ? y y0 v0 yt 12 a yt 2 2(00050 m) 640 1013 m/s2 t2 (125 108 s)2 (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 ay 2( y y0 ) E ma (9109 1031 kg)(640 1013 m/s 2 ) 364 N/C e 1602 1019 C 4 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 eE (1602 1019 C)(364 N/C) 349 1010 m/s 2 27 mp 1673 10 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 125 108 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 a y 349 1010 m/s2 y y0 v0 yt 12 a yt 2 12 (349 1010 m/s 2 )(125 108 s)2 273 106 m The displacement is 273 106 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. (d) In each case a g and it is reasonable to ignore the effects of gravity. 21.39.IDENTIFY: Find the angle that r̂ makes with the x-axis. Then rˆ (cos ) iˆ sin ˆj. . SET UP: tan y/x 135 ˆ EXECUTE: (a) tan 1 rad. rˆ j. 2 0 2ˆ 2 ˆ 12 i j. (b) tan 1 rad. rˆ 2 2 12 4 26 ˆ ˆ (c) tan 1 197 rad 1129 rˆ 039i 092 j (Second quadrant). 110 EVALUATE: In each case we can verify that r̂ is a unit vector, because rˆ rˆ 1. 21.45.IDENTIFY: Eq. (21.7) gives the electric field of each point charge. Use the principle of superposition and add the electric field vectors. In part (b) use Eq. (21.3) to calculate the force, using the electric field calculated in part (a). (a) SET UP: The placement of charges is sketched in Figure 21.45a. \ Figure 21.45a The electric field of a point charge is directed away from the point charge if the charge is positive and toward 1 q the point charge if the charge is negative. The magnitude of the electric field is E , where r is the 4e0 r 2 distance between the point where the field is calculated and the point charge. (i) At point a the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45b. Figure 21.45b EXECUTE: E1 9 1 q1 C 9 2 2 200 10 8 988 10 N m /C 4494 N/C 2 4 e0 r12 (0200 m) 5 E2 9 1 q2 C 9 2 2 500 10 (8 988 10 N m /C ) 1248 N/C 2 2 4 e0 r2 (0600 m) E1x 4494 N/C, E1 y 0 E2 x 1248 N/C, E2 y 0 Ex E1x E2 x 4494 N/C 1248 N/C 5742 N/C E y E1 y E2 y 0 The resultant field at point a has magnitude 574 N/C and is in the x-direction. (ii) SET UP: At point b the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45c. Figure 21.45c EXECUTE: E1 E2 1 q1 200 109 C 8988 109 N m 2/C 2 125 N/C 2 4 e0 r1 (120 m) 2 1 q2 500 109 C 8988 109 N m 2 /C 2 2809 N/C 2 4 e0 r2 (0400 m) 2 E1x 125 N/C, E1 y 0 E2 x 2809 N/C, E2 y 0 Ex E1x E2 x 125 N/C 2809 N/C 2684 N/C E y E1 y E2 y 0 The resultant field at point b has magnitude 268 N/C and is in the x-direction. (iii) SET UP: At point c the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45d. Figure 21.45d EXECUTE: E1 E2 9 1 q1 C 9 2 2 200 10 (8 988 10 N m /C ) 4494 N/C 2 2 4 e0 r1 (0200 m) 9 1 q2 C 9 2 2 500 10 (8 988 10 N m /C ) 449 N/C 2 4 e0 r22 (100 m) E1x 4494 N/C, E1 y 0 E2 x 449 N/C, E2 y 0 Ex E1x E2 x 4494 N/C 449 N/C 4045 N/C E y E1 y E2 y 0 The resultant field at point b has magnitude 404 N/C and is in the x-direction. (b) SET UP: Since we have calculated E at each point the simplest way to get the force is to use F eE EXECUTE: (i) F (1602 1019 C)(5742 N/C) 920 1017 N, x-direction (ii) F (1602 1019 C)(2684 N/C) 430 1017 N, x-direction (iii) F (1602 1019 C)(4045 N/C) 648 1017 N, x-direction 6 EVALUATE: The general rule for electric field direction is away from positive charge and toward negative charge. Whether the field is in the x- or x-direction depends on where the field point is relative to the charge that produces the field. In part (a), for (i) the field magnitudes were added because the fields were in the same direction and in (ii) and (iii) the field magnitudes were subtracted because the two fields were in opposite directions. In part (b) we could have used Coulomb’s law to find the forces on the electron due to the two charges and then added these force vectors, but using the resultant electric field is much easier. q 21.47.IDENTIFY: E k 2 . The net field is the vector sum of the fields due to each charge. r SET UP: The electric field of a negative charge is directed toward the charge. Label the charges q1, q2 and q3 , as shown in Figure 21.47a. This figure also shows additional distances and angles. The electric fields at point P are shown in Figure 21.47b. This figure also shows the xy coordinates we will use and the x and y components of the fields E1 , E 2 and E3. EXECUTE: E1 E3 (899 109 N m 2 /C2 ) E2 (899 109 N m 2 /C2 ) 200 106 C (00600 m) 2 500 106 C (0100 m) 2 449 106 N/C 499 106 N/C E y E1 y E2 y E3 y 0 and Ex E1x E2 x E3x E2 2E1 cos531 104 107 N/C E 104 107 N/C, toward the 200C charge. EVALUATE: The x-components of the fields of all three charges are in the same direction. Figure 21.47 21.52. IDENTIFY: For a long straight wire, E SET UP: . 1 180 1010 N m 2 /C2 . 2 e0 EXECUTE: r 21.54. 2 e0 r 15 1010 C/m 108 m 2e0 (250 N/C) EVALUATE: For a point charge, E is proportional to 1/r 2 For a long straight line of charge, E is proportional to 1/r (a) IDENTIFY: The field is caused by a finite uniformly charged wire. SET UP: The field for such a wire a distance x from its midpoint is E 1 1 2 . 2 2 e0 x ( x/a ) 1 4 e0 x ( x/a) 2 1 7 EXECUTE: E (180 109 N m2 /C2 )(175 109 C/m) 2 3.03 104 N/C, directed upward. 600 cm (00600 m) 1 425 cm (b) IDENTIFY: The field is caused by a uniformly charged circular wire. SET UP: The field for such a wire a distance x from its midpoint is E 1 Qx . We first find the 4e0 ( x2 a2 )3/2 radius of the circle using 2 r l. EXECUTE: Solving for r gives r l/2 (8.50 cm)/2 1.353 cm. The charge on this circle is Q l (175 nC/m)(0.0850 m) 14.88 nC. The electric field is E 1 Qx (900 109 N m2/C2 )(1488 102 9 C/m)(00600 m) 3/2 4 e0 ( x 2 a 2 )3/2 (00600 m)2 (001353 m)2 E = 3.45 104 N/C, upward. EVALUATE: In both cases, the fields are of the same order of magnitude, but the values are different because the charge has been bent into different shapes. 8