Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 25 - Electric Charges and Forces - Exercises and Problems - Page 747: 52

Answer

$\pm 8.2 \;\rm nC $

Work Step by Step

We know that the electric force between the two balls is an attractive force and its magnitude is given by Coulomb’s law. $$F=\dfrac{kq_1q_2}{r^2}$$ We are given that the magnitude of the two charges in both balls is the same, so $q_1=q_2=q$. And we also know that the separation distance here is $d$. Thus, $$F=\dfrac{kq^2}{d^2}\tag 1$$ From Newton's second law, we know that $$F_{net}=ma\tag 2$$ and since the surface is slippery, the only force that is exerted on one of the balls is the attractive electric force from the other ball. Hence, $$F_{net}=F$$ From (1) and (2), $$\dfrac{kq^2}{d^2}=ma$$ Thus, $$a=\dfrac{kq^2}{md^2}$$ $$a=\dfrac{kq^2}{m}\cdot \dfrac{1}{d^2} \tag 3$$ where we can assume that this is a straight-line formula $y=mx+b$ where $y=a$, $x=\dfrac{1}{d^2}$, $m={\rm solpe}=\dfrac{kq^2}{m}$, Hence, we can find the charge from the slope $${\rm solpe}=\dfrac{kq^2}{m}$$ $$ q=\sqrt{\dfrac{m\;{\rm solpe}}{k}}$$ Plugging the known; $$ \boxed{q=\sqrt{\dfrac{0.002\;{\rm solpe}}{8.99\times 10^9}}}$$ Now we need to use (3) to find the points to draw the best fit-line and find its slope. As we see below, the slope is about $3.02\times10^{-4}\;\rm m^3/s^2 $. Plug it into the boxed formula above, $$ q=\sqrt{\dfrac{0.002\; (3.02\times10^{-4})}{8.99\times 10^9}}=\color{red}{\bf \pm 8.2\times 10^{-9}}\;\rm C $$
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