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: 22.17

Answer

(a) $E = 6.50 \times 10^{5} \mathrm{N} / \mathrm{C}$ (b) $E = 7.20 \times 10^{4} \mathrm{N} / \mathrm{C}$

Work Step by Step

(a) At y = 0.2 m, both lines apply electric field in the positive direction of y-axis, so the net electric field due to both lines equals the summation of their electric field \begin{align} E&=\frac{1}{2 \pi \epsilon_{0}} \frac{\lambda_{1}}{r}+\frac{1}{2 \pi \epsilon_{0}} \frac{\lambda_{2}}{r}\\ & =\frac{1}{2 \pi \epsilon_{0} r}\left(\lambda_{1}+\lambda_{2}\right)\\ &=\left(\frac{18 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}{0.2 \mathrm{m}}\right)(4.8+2.4) 10^{-6} \mathrm{C} / \mathrm{m}\\ &= \boxed{6.5 \times 10^{5} \mathrm{N} / \mathrm{C}} \end{align} (b) At y = 0.6 m, both lines apply electric fields in opposite directions, so the net electric field for both kine at y = 0.6 m is calculated by \begin{aligned} E&=\frac{1}{2 \pi \epsilon_{0}}\left(\frac{\lambda_{2}}{r_{2}}-\frac{\lambda_{1}}{r_{1}}\right) \\ &=\left(18 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(\frac{2.4 \times 10^{-6} \mathrm{C} / \mathrm{m}}{0.2 \mathrm{m}}-\frac{4.8 \times 10^{-6} \mathrm{C} / \mathrm{m}}{0.6 \mathrm{m}}\right) \\ &=\boxed{7.2 \times 10^{4} \mathrm{N} / \mathrm{C}} \end{aligned}
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