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: 12


(a) New distance $r_{2}$ is less than 10 times of the original distance $r_{1}$ $ r_{2} =\sqrt10 r_{1}=3.162 r_{1}$ (b) new distance will be $ 94.868cm$

Work Step by Step

Let us assume two charges $q_{1}$ and $q_{2}$ are separated by a distance $r_{1}$. In this case magnitude of force between them will be $F_{1}=k\frac{q_{1}\times q_{2}}{r_{1}^2}$ ........equation (1) Now suppose the are now moved farther apart to a distance $r_{2}$,In this case suppose force between them is $F_{2}$ which is decreased by a factor of 10. $F_{2}=k\frac{q_{1}\times q_{2}}{r_{2}^{2}}$ ........equation (2) given that $F_{2}=\frac{1}{10}\times F_{1}$ putting values of $F_{2}$ and $F_{1}$ from equation(1) and (2) we will get $k\frac{q_{1}\times q_{2}}{r_{2}^{2}}=\frac{1}{10}\times k\frac{q_{1}\times q_{2}}{r_{1}^2}$ it gives $r_{2}^{2}=10\times r_{1}^{2}$ or $ r_{2} =\sqrt10 r_{1}=3.162 r_{1}$....equation(3) (b) original distance $ r_{1} =30cm$ so new distance will be $ r_{2} =3.162 r_{1}=3.162\times 30cm r_{1}=94.868cm$
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