Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 22 - Electric Charges and Forces - Exercises and Problems: 16

Answer

(a) $F = 57.6~N$ (b) $F = 4.65\times 10^{-35}~N$ (c) The ratio of the electric force to the gravitational force is $1.24\times 10^{36}$

Work Step by Step

(a) We can find the magnitude of the electric force: $F = \frac{k~q_1~q_2}{r^2}$ $F = \frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)(1.6\times 10^{-19}~C)}{(2.0\times 10^{-15}~m)^2}$ $F = 57.6~N$ (b) We can find the magnitude of the gravitational force: $F = \frac{G~m_1~m_2}{r^2}$ $F = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(1.67\times 10^{-27}~kg)(1.67\times 10^{-27}~kg)}{(2.0\times 10^{-15}~m)^2}$ $F = 4.65\times 10^{-35}~N$ (c) We can find the ratio of the electric force to the gravitational force: $\frac{57.6~N}{4.65\times 10^{-35}~N} = 1.24\times 10^{36}$
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