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 - Page 628: 76

Answer

$F = 1.25\times 10^{-4}~N$

Work Step by Step

We can find the distance $r$ between the $10~nC$ charge and the $-1.0~nC$ charge: $\frac{r}{5.0~cm} = cos~60^{\circ}$ $r = (5.0~cm)~cos~60^{\circ}$ $r = 2.5~cm$ We can find the magnitude of $F$: $F = \vert \frac{k~q_1~q_2}{r^2} \vert~sin~60^{\circ}$ $F = \frac{(9.0\times 10^9~N~m^2/kg^2)(1.0\times 10^{-9}~C)(10\times 10^{-9}~C)}{(0.025~m)^2}~sin~60^{\circ}$ $F = 1.25\times 10^{-4}~N$
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