College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 16 - Problems - Page 617: 75


$\lambda = -3.72\times 10^{-23}~C/m$

Work Step by Step

Since the electron is suspended, the electric force exerted on the electron is equal in magnitude to the electron's weight. We can find the magnitude of the linear charge density: $E~\vert q \vert = mg$ $(\frac{\lambda}{2\pi~\epsilon_0~r})~(\vert q \vert) = mg$ $\lambda = \frac{2\pi~\epsilon_0~r~mg}{\vert q \vert}$ $\lambda = \frac{(2\pi)~(8.854\times 10^{-12}~F/m)~(0.012~m)~(9.109\times 10^{-31}~kg)(9.80~m/s^2)}{1.6\times 10^{-19}~C}$ $\lambda = 3.72\times 10^{-23}~C/m$ Since the electric force exerted on the electron is directed upward, the linear charge density must have a negative value. Therefore, $\lambda = -3.72\times 10^{-23}~C/m$
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