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Physics 51 Chapter 21 (parts A and B), 13th ed, Freedman Homework Solutions 21.9. IDENTIFY: Apply Coulomb’s law. SET UP: Consider the force on one of the spheres. (a) EXECUTE: q1 q2 q F 1 q1q2 q2 so 4 e0 r 2 4 e0 r 2 qr F 0220 N 0150 m 742 107 C (on each) 9 2 2 (1/4 e0 ) 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 2 4 e0 r 4(1/4 e0 ) 2 (1/4 e0 ) 2 4 e0 r 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 21.16. 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 . IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on Q. SET UP: The force that q1 exerts on Q is repulsive, as in Example 21.4, but now the force that q2 exerts is attractive. EXECUTE: The x-components cancel. We only need the y-components, and each charge contributes 1 (20 106 C)(40 106 C) sin 0173 N (since sin 0600) equally. F1 y F2 y 4 e0 (0500 m) 2 Therefore, the total force is 2F 035 N, in the y-direction. EVALUATE: If q1 is 20C and q2 is 20C, then the net force is in the y-direction. 21.19. IDENTIFY and SET UP: Apply Coulomb’s law to calculate the force exerted by q2 and q3 on q1 Add these forces as vectors to get the net force. The target variable is the x-coordinate of q3 EXECUTE: F2 is in the x-direction. F2 k q1q2 r122 337 N, so F2 x 337 N Fx F2 x F3x and Fx 700 N F3x Fx F2 x 700 N 337 N 1037 N For F3x to be negative, q3 must be on the x-axis. F3 k q1q3 x 2 , so x k q1q3 F3 0144 m, so x 0144 m EVALUATE: q2 attracts q1 in the x-direction so q3 must attract q1 in the x-direction, and q3 is at negative x. 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. 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 EXECUTE: q FE w so eE mg mg (1673 1027 kg)(980 m/s 2 ) 102 107 N/C 19 e 1602 10 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 EXECUTE: E ( 108 ) than in part (a) ( 102 ) so E is much smaller in (b). For subatomic particles gravity can usually be ignored compared to electric forces. 21.35. IDENTIFY: Apply constant acceleration equations to the motion of the electron. SET UP: Let x be to the right and let y be downward. The electron moves 2.00 cm to the right and 0.50 cm downward. EXECUTE: Use the horizontal motion to find the time when the electron emerges from the field. x x0 00200 m, ax 0, v0 x 160 106 m/s. x x0 v0 xt 12 axt 2 gives t 125 108 s. Since v0 y v y ax 0, vx 160 106 m/s. y y0 00050 m, v0y 0, t 125 108 s. y y0 t gives 2 v y 800 105 m/s. Then v vx2 v 2y 179 106 m/s. EVALUATE: v y v0 y a yt gives a y 64 1013 m/s2. The electric field between the plates is E 21.46. ma y e (911 1031 kg)(64 1013 m/s2 ) 160 1019 C 364 V/m. This is not a very large field. IDENTIFY: Apply Eq. (21.7) to calculate the field due to each charge and then require that the vector sum of the two fields to be zero. SET UP: The field of each charge is directed toward the charge if it is negative and away from the charge if it is positive. EXECUTE: The point where the two fields cancel each other will have to be closer to the negative charge, because it is smaller. Also, it can’t be between the two charges, 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 400 nC charge. The zero point will therefore have to be a factor of 21.47. 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: 120 x 2 x and x 290 m. EVALUATE: This point is 4.10 m from the 8.00 nC charge. The two fields at this point are in opposite directions and have equal magnitudes. q 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 /C 2 ) 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: 2 e0 r . 1 180 1010 N m2 /C2 . 2e0 EXECUTE: 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 21.54. (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 EXECUTE: E 1 1 2 . 2 e0 x ( x/a)2 1 4 e0 x ( x/a) 2 1 (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 2 4e0 ( x a2 )3/2 find the 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) 2 2 3/2 3/2 4 e0 ( x a ) (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. 21.73. IDENTIFY: The electric field exerts a horizontal force away from the wall on the ball. When the ball hangs at rest, the forces on it (gravity, the tension in the string, and the electric force due to the field) add to zero. SET UP: The ball is in equilibrium, so for it Fx 0 and Fy 0. The force diagram for the ball is given in Figure 21.73. FE is the force exerted by the electric field. F qE . Since the electric field is horizontal, FE is horizontal. Use the coordinates shown in the figure. The tension in the string has been replaced by its x- and y-components. Figure 21.73 EXECUTE: Fy 0 gives Ty mg 0. T cos mg 0 and T mg . Fx 0 gives cos FE Tx 0. FE T sin 0. Combing the equations and solving for FE gives mg 3 2 2 FE sin mg tan (12.3 10 kg)(9.80 m/s )(tan17.4) 3.78 10 N. FE q E so cos FE 3.78 102 N 3.41 104 N/C. Since q is negative and FE is to the right, E is to the left in the q 1.11 106 C figure. E EVALUATE: The larger the electric field E the greater the angle the string makes with the wall. 21.89. IDENTIFY: Divide the charge distribution into infinitesimal segments of length dx. Calculate E x and E y due to a segment and integrate to find the total field. SET UP: The charge dQ of a segment of length dx is dQ (Q/a)dx. The distance between a segment at x and the charge q is a r x. (1 y)1 1 y when y EXECUTE: (a) dEx a r x, so Ex (b) F qE 1. 1 a Qdx 1 Q1 1 1 dQ so Ex . 2 2 0 4 e0 a(a r x) 4 e0 a r a r 4 e0 (a r x) 1 Q 1 1 . E y 0. 4 e0 a x a x 1 qQ 1 1 ˆ i . 4 e0 a r a r kqQ kqQ kqQ 1 qQ ((1 a/x)1 1) (1 a/x 1) 2 . (Note ax ax 4 e0 r 2 x that for x a, r x a x.) The charge distribution looks like a point charge from far away, so the force takes the form of the force between a pair of point charges. EVALUATE: (c) For x 21.103. a, F IDENTIFY and SET UP: Example 21.11 gives the electric field due to one infinite sheet. Add the two fields as vectors. EXECUTE: The electric field due to the first sheet, which is in the xy-plane, is E ( /2e )kˆ for z 0 1 0 and E1 ( /2e0 )kˆ for z 0 We can write this as E1 /2e0 z/ z kˆ, since z/ z 1 for z 0 and z/ z z/z 1 for z 0 Similarly, we can write the electric field due to the second sheet as E2 ( /2e0 )( x/ x )iˆ, since its charge density is The net field is E E E ( /2e )( x/ x iˆ ( z/ z )kˆ) 1 2 0 EVALUATE: The electric field is independent of the y-component of the field point since displacement in the y-direction is parallel to both planes. The field depends on which side of each plane the field is located.