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: 48

Answer

$\alpha = 4$

Work Step by Step

Let $F_x$ be the magnitude of the force on the charge $-q$ (directed to the right) from the charge $\alpha q$. Let $F_y$ be the magnitude of the force on the charge $-q$ (directed upward) from the charge $q$. Since the net force is directed at a $45^{\circ}$ angle, then $F_x = F_y$. We can find $\alpha$: $F_x = F_y$ $\frac{k~q~\alpha q}{(2L)^2} = \frac{k~q~q}{L^2}$ $\frac{\alpha}{4} = 1$ $\alpha = 4$
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