Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 25 - Electric Charges and Forces - Exercises and Problems: 21

Answer

(a) $F = (6.4\times 10^{-17}~\hat{i}+~1.6\times 10^{-17}~\hat{j})~N$ (b) $F = (-6.4\times 10^{-17}~\hat{i}-~1.6\times 10^{-17}~\hat{j})~N$ (c) $a = 4.0\times 10^{10}~m/s^2$ (d) $a = 7.3\times 10^{13}~m/s^2$

Work Step by Step

(a) We can find the force on the proton. $F = E~q$ $F = [(400~\hat{i}+~100~\hat{j})~N/C]~(1.6\times 10^{-19}~C)$ $F = (6.4\times 10^{-17}~\hat{i}+~1.6\times 10^{-17}~\hat{j})~N$ (b) We can find the force on the electron. $F = E~q$ $F = [(400~\hat{i}+~100~\hat{j})~N/C]~(-1.6\times 10^{-19}~C)$ $F = (-6.4\times 10^{-17}~\hat{i}-~1.6\times 10^{-17}~\hat{j})~N$ (c) We can find the magnitude of the force. $F = \sqrt{F_x^2+F_y^2}$ $F = \sqrt{(6.4\times 10^{-17}~N)^2+(1.6\times 10^{-17}~N)^2}$ $F = 6.6\times 10^{-17}~N$ We can find the magnitude of the proton's acceleration. $a = \frac{F}{m}$ $a = \frac{6.6\times 10^{-17}~N}{1.67\times 10^{-27}~kg}$ $a = 4.0\times 10^{10}~m/s^2$ (d) We can find the magnitude of the force. $F = \sqrt{F_x^2+F_y^2}$ $F = \sqrt{(-6.4\times 10^{-17}~N)^2+(-1.6\times 10^{-17}~N)^2}$ $F = 6.6\times 10^{-17}~N$ We can find the magnitude of the electron's acceleration. $a = \frac{F}{m}$ $a = \frac{6.6\times 10^{-17}~N}{9.1\times 10^{-31}~kg}$ $a = 7.3\times 10^{13}~m/s^2$
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