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 - Exercises - Page 556: 13


Initial separation was $223.606cm$

Work Step by Step

Let us assume initially two charges $q_{1}$ and $q_{2}$ are separated by a distance $r_{1}$. In this case magnitude of force between them was $F_{1}=k\frac{q_{1}\times q_{2}}{r_{1}^2}$ ........equation (1) Now suppose they are brought together to a distance $r_{2}=100cm$, apart ,In this case suppose force between them is $F_{2}$ which is increased by a factor of 5 from the initial scenario. $F_{2}=k\frac{q_{1}\times q_{2}}{r_{2}^{2}}$ ........equation (2) given that $F_{2}=5\times F_{1}$ putting values of $F_{2}$ and $F_{1}$ from equation(1) and (2) we will get $k\frac{q_{1}\times q_{2}}{r_{2}^{2}}=5\times k\frac{q_{1}\times q_{2}}{r_{1}^2}$ it gives , $r_{1}^2=5\times r_{2}^2$ or $ r_{1} =\sqrt5 r_{2}$ since $ r_{2} =100cm$ $ r_{1}=\sqrt5 \times 100cm=223.606cm$
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