MasterPhysics 8 soln
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MasterPhysics 8 soln

Course Number: PHYS 408, Spring 2009

College/University: New Hampshire

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PHYS-408 Fall 2008 Mastering Physics Assignment #8 Solutions Problem 26.35 Two 1.0 g spheres are charged equally and placed 2.0 cm apart. When released, they begin to accelerate at 150 m/s2. (a) What is the magnitude of the charge on each sphere? From the acceleration, we can find the force r r F = ma F = (0.001kg) 150m /s2 = 0.150N ( ) This must equal the electrostatic force 1 q1 q2 1 q2 F= = 4 0 r 2 4 0 r 2...

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Fall PHYS-408 2008 Mastering Physics Assignment #8 Solutions Problem 26.35 Two 1.0 g spheres are charged equally and placed 2.0 cm apart. When released, they begin to accelerate at 150 m/s2. (a) What is the magnitude of the charge on each sphere? From the acceleration, we can find the force r r F = ma F = (0.001kg) 150m /s2 = 0.150N ( ) This must equal the electrostatic force 1 q1 q2 1 q2 F= = 4 0 r 2 4 0 r 2 q 2 = 4 0 r 2 F = 4 8.85 10-12 C 2 /Nm 2 (0.02m) (0.150N ) ( ) 2 = 6.67 10-15 C 2 q = 6.67 10-15 C 2 = 8.17 10-8 C = 81.7nC Problem 26.17 (a) What is the magnitude of the net electric force on charge A in the figure? The charge A is repelled by B with a magnitude -1.0 10-9 C -2.0 10-9 C 1 q1 q2 9 2 2 F1 = = 8.99 10 Nm /C 4 0 r 2 (0.02m) 2 ( ) = 4.50 10-5 N The charge A is attracted by C with a magnitude So, -1.0 10-9 C 2.0 10-9 C 1 q1 q2 9 2 2 F2 = = 8.99 10 Nm /C 4 0 r 2 (0.03m) 2 ( ) = 2.00 10-5 N F = F1 - F2 = 4.50 10-5 N - 2.00 10-5 N = 2.5 10-5 N (b) What is the direction of the net electric force on charge A in the figure? The force away is greater, so the direction is up. Problem 26.20 (a) What is the strength of the electric field E_p1 1.0 mm from a proton? r 1 q 1.60 10-19 C ^ ^ E= r = 8.99 10 9 Nm 2 /C 2 r 4 0 r 2 -3 2 10 m ( ) ( ) ^ = 1.44 10-19 N /C r ( ) (b) In which direction does the electric field point? Since r is away, so is the force. ^ (c) What is the strength of the electric field E_e1 1.0 mm from an electron? The magnitude of the charge of the electron is the same as the proton so the magnitude of the force is the same, 1.44 10-19 N /C . (d) In which direction does the electric field point? The charge of electron has the opposite sign as the charge on the proton, so the direction of the field is opposite, that is, towards the electron. Problem 26.33 Two protons are 2.0 fm apart. (a) What is the magnitude of the electric force on one proton due to the other proton? 1 FE = 4 0 q2 1.6 10-19 C ) 9 2 2 ( = (8.99 10 Nm /C ) 2 r2 (2.0 10-15 m) 2 = 57.6N (b) What is the magnitude of the gravitational force on one proton due to the other proton? 1.67 10-27 C ) m -11 2 2 ( FG = G 2 = (6.67 10 Nm /kg ) 2 r (2.0 m) 2 10-15 2 = 4.65 10-35 N (c) What is the ratio of the electric force to the gravitational force? FE 57.6N = = 1.34 10 36 Fg 4.65 10-35 N Problem 26.42 (a) What is the force F_vec on the 5.0 nC charge in the figure? Give your answer as a magnitude and an angle measured ccw from the +x-axis. The 5.0 nC charge experiences two forces. One, from the -10 nC charge: r F1 = 1 4 0 q1q2 2 r21 ^ r21 = 8.99 10 Nm ( 9 2 (5 10-9 C)(-10 10-9 C) r^21 /C ) 2 (4.0 10-2 m) 2 ^ = -2.8110-4 N r21 ^ The unit vector r12 is just ^ ^ r21 = - j r F1 = 2.8110-4 N ^ j The second is from the +10 nC charge: 5 10-9 C 10 10-9 C r 1 q1q2 9 2 2 ^ ^ F2 = r = 8.99 10 Nm /C r31 2 31 4 0 r31 -2 2 -2 2 4.0 10 m + 3.0 10 m ( ) ( ) ( ) ( )( ) ( ) ( ) ^ = 1.80 10-4 N r31 ( ) The radial unit vector here is just 3.0 10-2 m ^ r31 = - 2 (4.0 10-2 m) + (3.0 10-2 m) ( ) 2 ^ i- 4.0 10-2 m (4.0 10-2 m) + (3.0 10-2 m) 2 2 ^ j ^ = -0.6i - 0.8 ^ j So r ^ F2 = +1.80 10-4 N 0.6i + 0.8 ^ j Then r r r ^ F = F1 + F2 = 2.8110-4 N ^ - 1.80 10-4 N 0.6i + 0.8 ^ j j ( ) ( )( ) ^ = -1.08 10-4 N i + 1.37 10-4 N ^ j So, the magnitude of the force is just F= ( ) ( 2 ) (-1.08 10-4 N ) + (1.37 10-4 N ) 2 = 1.74 10-4 N The angle of the force is then F -1.08 10-4 N cos = x = = -0.621 F 1.74 10-4 N = arccos(-0.621) = 128o Problem 26.63 (a) What are the electric fields at points 1, 2, and 3 in the figure? Give your answer in component form. r E1 = 1 4 0 q r12 ^ r1 = 8.99 10 Nm ( 9 2 (5 10-9 C) ^ /C ) r1 -2 2 -2 2 (1.0 10 m) + (2.0 10 m) 2 ^ = 8.99 10 4 N /C r1 ( ) Then ^ r1 = 1.0 10-2 m (1.0 10-2 m) + (2.0 10-2 m) ( ) ( 2 2 ^ i+ 2.0 10-2 m (1.0 10-2 m) + (2.0 10-2 m) ) 2 2 ^ j ^ = 0.447i + 0.894 ^ j So r ^ E1 = 4.02 10 4 N /C i + 8.04 10 4 N /C ^ j (b) r E2 = 5 10-9 C 1 q 9 2 2 ^ ^ r = 8.99 10 Nm /C r2 2 1 4 0 r2 -2 2 1.0 10 m ( ) ( ( ) ) ^ = 4.5 10 5 N /C r2 ^ ^ But, r2 = i , so r ^ E 2 = 4.5 10 5 N /C i ( ) ( ) (c) This has the same magnitude as (a) r ^ E 3 = 8.99 10 4 N /C r3 The direction is the same as (a) except the ^ is negative j ^ - 0.894 ^ ^ r3 = 0.447i j So r ^ E 3 = 4.02 10 4 N /C i -8.04 10 4 N /C ^ j ( ) ( ) ( )

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