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: 16


(a) $q_{1}=1.08\times10^{-9}C$, $q_{2}=3.24\times10^{-9}C$ (b) $F\approx4.2\times10^{-6}N=4.2\mu N$

Work Step by Step

Let us assume two negative charges $q_{1}$ and $q_{2}$ are separated by a distance $r$ given that $r =10.0cm=10.0\times10^(-2)m=1.0\times10^{-1}m$ given $q_{2}=3\times q_{1}$ Repulsive force force between the charges is $F=3.15\mu N=3.15\times10^{-6}N$ (a) calculation of charges $F=k\frac{q_{1}\times q_{2}}{r_{}^2}$ putting $F=3.15\times10^{-6}N$, $r =1.0\times10^{-1}m$, ,$q_{2}=3\times q_{1}$ , $k=9.0\times10^{9}N.m^2/C^2$ $3.15\times10^{-6}N=9.0\times10^{9}N.m^2/C^2\times \frac{q_{1}\times3\times q_{1} }{(1.0\times10^{-1}m)^2}$ $3 q_{1}^2= 0.35\times10^{-17}C^2$ $q_{1}^2=0.11667\times10^{-17}C^2=1.1667\times10^{-18}C^2$ as the charges are negative $q_{1}=-1.08\times10^{-9}C$, other charge will be $q_{2}=3\times q_{2}=3\times-1.08\times10^{-9}C=-3.24\times10^{-9}C$ (b) total charge $q=q_{1}+q_{2}=-1.08\times10^{-9}C+(-3.24\times10^{-9})C=-4.32\times10^{-9}C$ if charges are equally distributed $q_{1}=q_{2}=\frac{q}{2}$=$\frac{-4.32\times10^{-9}C}{2}=-2.16\times10^{-9}C$ $F=k\frac{q_{1}\times q_{2}}{r_{}^2}$ putting $F=3.15\times10^{-6}N$, $r =1.0\times10^{-1}m$, ,$q_{2}= q_{1}=-2.16\times10^{-9}C$ , $k=9.0\times10^{9}N.m^2/C^2$ $F=9.0\times10^{9}N.m^2/C^2\times\frac{2.16\times10^{-9}C\times2.16\times10^{-9}C}{(1.0\times10^{-1}m)^2}$ $F=4.199\times10^{-6}N\approx4.2\times10^{-6}N$
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