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: 22.13

Answer

$F = 0.081 \mathrm{N}$

Work Step by Step

We will calacute the force due to the applied electric field on the charge where the charge $q$ is equal $$q=\lambda X = (5.2 \times 10^{-6} \mathrm{C} / \mathrm{m})(0.0500 \mathrm{m})=2.6 \times 10^{-7} \mathrm{C}$$ \begin{align*} F&=\frac{1}{2 \pi \epsilon_{o}} \frac{\lambda}{r} (q) \\ &=\frac{5.2 \times 10^{-6} \mathrm{C} / \mathrm{m}}{2 \pi\left(8.854 \times 10^{-12} \mathrm{C}^{2} /\mathrm{N} \cdot \mathrm{m}^{2}\right)(0.3 \mathrm{m})} (2.6 \times 10^{-7} \mathrm{C})\\ &= \boxed{0.081 \mathrm{N}} \end{align*}
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