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: 22.10

Answer

(a) $\Phi_{E} = 0 $ (b) $\Phi_{E} = -678 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$ (c) $\Phi_{E} = -226 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$

Work Step by Step

(a) Inside the conductor, there are no excess charges so the electric flux is zero $$ \Phi_{E}=0 $$ (b) Inside radius 1.5 cm, the sphere contains $q_2$ only, so the electric flux will be $$ \Phi_{E}=\frac{q_{2}}{\epsilon_{0}}=\frac{-6.00 \times 10^{-9} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}=-678 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} $$ (c) Inside radius 2.5 cm, the sphere contains charges $q_1$ and $q_2$ and the electric flux will be $$ \Phi_{E}=\frac{q_{1}+q_{2}}{\epsilon_{0}}=\frac{(4.00-6.00) \times 10^{-9} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}=-226 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} $$
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