Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 683: 13

Answer

Greater force is experienced in the center of the square (see step-by-step for details).

Work Step by Step

The magnitude of the force between two point charges, $q_{1}$ and $q_{2}$, separated by a distance $r$ is $F =k\displaystyle \frac{|q_{1}||q_{2}|}{r^{2}},\qquad (19- 5)$ (Force is inversely proportional to the square of the distance between charges.) Let the square have sides a. Then the diagonal is $d=a\sqrt{2}$. The center of the square is at a distance of about $0.7a$ from the corners. Let the third charge be positive. Placing the third charge in the center, two forces will act on it, along the diagonal one repelling from the +q charge, acting towards the -q charge, and the other attracting, toward the -q charge. The net force has magnitude equal to the sum of these two, because they are colinear. Placing the third charge in a corner, it would be at a greater distance from either charge, so the forces acting would be weaker than above. Furthermore, their directions are perpendicular, so the net force would have magnitude less than their sum of magnitudes, since they are not colinear. Conclusion: greater force is experienced in the center of the square.
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