College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 15 - Electric Charge, Forces, and Fields - Learning Path Questions and Exercises - Conceptual Questions - Page 555: 10

Answer

The distance to be increased to 3 times its initial distance.

Work Step by Step

Let $F=k\frac{q_{1}q_{2}}{d^{2}}$, When one charge is increased by 27 times, and other reduce to one-third, then new $F=k\frac{27q_{1}\frac{1}{3}q_{2}}{(d')^{2}}=9k\frac{q_{1}q_{2}}{(d^{'})^{2}}=k\frac{q_{1}q_{2}}{d^{2}}$, So, $d^{'}=3d$ The distance to be increased to 3 times its initial distance.
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