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: 19

Answer

(a) The acceleration of the proton is $1.7\times 10^{14}~m/s^2$ away from the glass bead. (b) The acceleration of the electron is $3.2\times 10^{17}~m/s^2$ toward the glass bead.

Work Step by Step

(a) We can find the magnitude of the electric force on the proton: $F = \frac{k~q_b~q_p}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)(20\times 10^{-9}~C)(1.6\times 10^{-19}~C)}{(0.010~m)^2}$ $F = 2.88\times 10^{-13}~N$ We can find the magnitude of the acceleration. $a = \frac{F}{m}$ $a = \frac{2.88\times 10^{-13}~N}{1.67\times 10^{-27}~kg}$ $a = 1.7\times 10^{14}~m/s^2$ Since both charges are positive, the force on the proton is directed away from the glass bead. The acceleration of the proton is $1.7\times 10^{14}~m/s^2$ away from the glass bead. (b) We can find the magnitude of the electric force on the electron: $F = \frac{k~q_b~q_p}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)(20\times 10^{-9}~C)(1.6\times 10^{-19}~C)}{(0.010~m)^2}$ $F = 2.88\times 10^{-13}~N$ We can find the magnitude of the acceleration. $a = \frac{F}{m}$ $a = \frac{2.88\times 10^{-13}~N}{9.1\times 10^{-31}~kg}$ $a = 3.2\times 10^{17}~m/s^2$ Since the glass bead's charge is positive and the electron's charge is negative, the force on the electron is directed toward the glass bead. The acceleration of the electron is $3.2\times 10^{17}~m/s^2$ toward the glass bead.
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