University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 22 - Gauss's Law - Problems - Exercises - Page 746: 22.15

Answer

$N = 1.35 \times 10^{10} \mathrm{electron}$

Work Step by Step

First, we will find the total charge $q$ where the electric field is related to the charge $q$ by \begin{gather} E=\frac{1}{4 \pi \epsilon_{\odot}} \frac{q}{r^{2}}\\ q = 4 \pi \epsilon_{\circ} r^{2} E \\ q =\frac{(1150 \mathrm{N} / \mathrm{C})}{9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}\left(13 \times 10^{-2} \mathrm{m}\right)^{2}\\ q =2.16 \times 10^{-9} \mathrm{C} \end{gather} Now, we divide this charge by the charge of electrons to get the number of the excess electrons \begin{aligned} N &=\frac{q}{e} \\ &=\frac{\left(2.16 \times 10^{-9} \mathrm{C}\right)}{1.6 \times 10^{-19} \mathrm{C} / \mathrm{electron}} \\ &=1.35 \times 10^{10} \mathrm{electron} \end{aligned}
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