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 26 - The Electric Field - Exercises and Problems - Page 775: 10

Answer

$+40\;\rm nC$

Work Step by Step

We know that the electric field exerted on a point at distance $r$ from the center of a long charged rod is given by $$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+\left(\frac{L}{2}\right)^2}}$$ Thus the repulsive force between the bead and the rod at $r$ distance is given by $$F=|q|E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|q||Q|}{r\sqrt{r^2+\left(\frac{L}{2}\right)^2}}$$ Hence, the total charge in the rod is given by $$|Q|=\dfrac{4\pi \epsilon_0 rF\;\sqrt{r^2+\left(\frac{L}{2}\right)^2} }{|q|}$$ Plugging the known; $$|Q|=\dfrac{4\pi (8.85\times 10^{-12}) (0.04)(840\times 10^{-6})\;\sqrt{0.04^2+\left(\frac{0.10}{2}\right)^2} }{|6\times 10^{-9}|}=\bf\pm 40\;\rm nC$$ And since the force between the bead and the rod is a repulsive force, then the charge of the rod must be positive. $$Q=\color{red}{\bf +40}\;\rm nC$$
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