Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 22 - Electric Fields - Problems - Page 656: 47a

Answer

$a = 1.92\times 10^{12}~m/s^2$

Work Step by Step

We can find the force on the proton: $F = q~E$ $F = (1.6\times 10^{-19}~C)(2.00\times 10^4~N/C)$ $F = 3.2\times 10^{-15}~N$ We can find the acceleration: $F = ma$ $a = \frac{F}{m}$ $a = \frac{3.2\times 10^{-15}~N}{1.67\times 10^{-27}~kg}$ $a = 1.92\times 10^{12}~m/s^2$
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