College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 16 - Problems - Page 618: 94

Answer

The ratio of $F_{ca}$ to $F_{ba}$ is $\frac{1}{2}$

Work Step by Step

Let $L$ be the length of each side of the square. We can find an expression for $F_{ba}$: $F_{ba} = \frac{k~Q^2}{L^2}$ We can find an expression for $F_{ca}$: $F_{ca} = \frac{k~Q^2}{(\sqrt{2}~L)^2} = \frac{k~Q^2}{2~L^2}$ We can find the ratio of $F_{ca}$ to $F_{ba}$: $\frac{F_{ca}}{F_{ba}} = \frac{\frac{k~Q^2}{2~L^2}}{\frac{k~Q^2}{L^2}} = \frac{1}{2}$
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