17_InstSolManual_PDF_Part5

17_InstSolManual_PDF_Part5 - Figure 17.20 y F 2 F 1 x 1...

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Solve: proton: electron: The proton has an initial acceleration of toward the electron and the electron has an initial accelera- tion of toward the proton. Note that the force the electron exerts on the proton is equal in magnitude to the force the proton exerts on the electron, but the accelerations produced by this force are different because the masses are different. 17.19. Set Up: with An electron has charge Solve: The spheres have equal charges q, so and The charges on the spheres have the same sign so the electri- cal force is repulsive and the spheres accelerate away from each other. Reflect: As the spheres move apart the repulsive force they exert on each other decreases and their acceleration decreases. 17.20. Set Up: The net force is the vector sum of the forces exerted by each electron. Each force is attractive so is directed toward the electron that exerts it. Solve: Each force has magnitude The vector force diagram is shown in Figure 17.20.
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Unformatted text preview: Figure 17.20 y F 2 F 1 x 1 65.0 ° F 1 5 F 2 5 k q 1 q 2 r 2 5 k e 2 r 2 5 1 8.988 3 10 9 N # m 2 / C 2 21 1.60 3 10 2 19 C 2 2 1 1.50 3 10 2 10 m 2 2 5 1.023 3 10 2 8 N. q p 5 1 1.60 3 10 2 19 C. q e 5 2 1.60 3 10 2 19 C. N 5 q e 5 2.29 3 10 2 6 C 1.60 3 10 2 19 C 5 1.43 3 10 13 electrons. q 5 r Å F k 5 1 0.150 m 2 Å 2.09 N 8.99 3 10 9 N # m 2 / C 2 5 2.29 3 10 2 6 C. F 5 k q 2 r 2 F 5 ma 5 1 8.55 3 10 2 3 kg 21 245 m / s 2 2 5 2.09 N. 2 e 5 2 1.60 3 10 2 19 C. F 5 k q 1 q 2 r 2 . F 5 ma , a 5 25.0 g 5 245 m / s 2 . 6.3 3 10 21 m / s 2 3.4 3 10 18 m / s 2 a e 5 F m e 5 5.75 3 10 2 9 N 9.11 3 10 2 31 kg 5 6.3 3 10 21 m / s 2 a p 5 F m p 5 5.75 3 10 2 9 N 1.67 3 10 2 27 kg 5 3.4 3 10 18 m / s 2 F 5 k e 2 r 2 5 1 8.99 3 10 9 N # m 2 / C 2 2 1 1.60 3 10 2 19 C 2 2 1 2.0 3 10 2 10 m 2 2 5 5.75 3 10 2 9 N Electric Charge and Electric Field 17-5...
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