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: 45

Answer

$F_x = \frac{-2~a~k~q~Q}{(a^2+y^2)^{3/2}}$

Work Step by Step

We can find the distance $r$ from the charge $q$ to each of the other charges. $r = \sqrt{a^2+y^2}$ By symmetry, the vertical component of each force on the charge $q$ cancels out. The net force on the charge $q$ is the sum of the horizontal component of each force. We can find the magnitude of the horizontal component of the force on the charge $q$. $F = 2\times \frac{k~q~Q}{r^2}~cos(\theta)$ $F = 2\times \frac{k~q~Q}{(\sqrt{a^2+y^2})^2}~(\frac{a}{\sqrt{a^2+y^2}})$ $F = \frac{2~a~k~q~Q}{(a^2+y^2)^{3/2}}$ Since the x-component of the force is directed to the left, the x-component of the force is negative. $F_x = \frac{-2~a~k~q~Q}{(a^2+y^2)^{3/2}}$
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