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Chapter 21 HW Solutions

Chapter 21 HW Solutions - PHYS 241 21.9 Ch 21 Homework...

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PHYS 241 Ch. 21 Homework Solutions 21.9. I DENTIFY : Apply F ma = , with 1 2 2 q q F k r = . S ET U P : 2 25.0 245 m/s a g = = . An electron has charge 19 1.60 10 C. e - - = - × E XECUTE : 3 2 (8.55 10 kg)(245 m/s ) 2.09 N F ma - = = × = . The spheres have equal charges q , so 2 2 q F k r = and 6 9 2 2 2.09 N (0.150 m) 2.29 10 C 8.99 10 N m /C F q r k - = = = × × . 6 13 19 2.29 10 C 1.43 10 electrons 1.60 10 C q N e - - × = = = × × . The charges on the spheres have the same sign so the electrical force is repulsive and the spheres accelerate away from each other. E VALUATE : As the spheres move apart the repulsive force they exert on each other decreases and their acceleration decreases. 21.17. I DENTIFY and S ET U P : Apply Coulomb’s law to calculate the force exerted by 2 q and 3 q on 1 . q Add these forces as vectors to get the net force. The target variable is the x -coordinate of 3 . q E XECUTE : 2 F r is in the x -direction. 1 2 2 2 2 12 3.37 N, so 3.37 N x q q F k F r = = = + 2 3 and 7.00 N x x x x F F F F = + = - 3 2 7.00 N 3.37 N 10.37 N x x x F F F = - = - - = - For 3 x F to be negative, 3 q must be on the x - -axis. 1 3 1 3 3 2 3 , so 0.144 m, so 0.144 m q q k q q F k x x x F = = = = - E VALUATE : 2 q attracts 1 q in the x + -direction so 3 q must attract 1 q in the x - -direction, and 3 q is at negative x. 21.23. I DENTIFY : Apply Coulomb’s law to calculate the force exerted on one of the charges by each of the other three and then add these forces as vectors. (a) S ET U P : The charges are placed as shown in Figure 21.23a. 1 2 3 4 q q q q q = = = = Figure 21.23a Consider forces on 4 . q The free-body diagram is given in Figure 21.23b. Take the y -axis to be parallel to the diagonal between 2 q and 4 q and let y + be in the direction away from 2 . q Then 2 F r is in the y + -direction. 1
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E XECUTE : 2 3 1 2 0 1 4 q F F L π = = P 2 2 2 0 1 4 2 q F L π = P 1 1 1 sin45 / 2 x F F F = - ° = - 1 1 1 cos45 / 2 y F F F = + ° = + 3 3 3 sin45 / 2 x F F F = + ° = + 3 3 3 cos45 / 2 y F F F = + ° = + 2 2 2 0, x y F F F = = Figure 21.23b (b) 1 2 3 0 x x x x R F F F = + + = 2 2 2 1 2 3 2 2 2 0 0 0 1 1 (2/ 2) (1 2 2) 4 4 2 8 y y y y q q q R F F F L L L π π π = + + = + = + P P P 2 2 0 (1 2 2). 8 q R L π = + P Same for all four charges. E VALUATE : In general the resultant force on one of the charges is directed away from the opposite corner. The forces are all repulsive since the charges are all the same. By symmetry the net force on one charge can have no component perpendicular to the diagonal of the square. 21.27. I DENTIFY : The acceleration that stops the charge is produced by the force that the electric field exerts on it. Since the field and the acceleration are constant, we can use the standard kinematics formulas to find acceleration and time. (a) S ET U P : First use kinematics to find the proton’s acceleration. 0 x v = when it stops. Then find the electric field needed to cause this acceleration using the fact that F = qE . E XECUTE : 2 2 0 0 2 ( ) x x x v v a x x = + - . 0 = (4.50 × 10 6 m/s) 2 + 2 a (0.0320 m) and a = 3.16 × 10 14 m/s 2 . Now find the electric field, with q = e . eE = ma and E = ma/e = (1.67 × 27 10 - kg)(3.16 × 10 14 m/s 2 )/(1.60 × 19 10 - C) = 3.30 × 10 6 N/C, to the left.
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