University Physics with Modern Physics (14th Edition)

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

Chapter 23 - Electric Potential - Problems - Exercises - Page 781: 23.62

Answer

$V = 1157 \mathrm{V}$

Work Step by Step

The potential difference of the cylinder shape is calculated by \begin{align} V&=Er \ln (\frac{b}{a}) \\ &= \left(2.00 \times 10^{4} \mathrm{N} / \mathrm{C}\right)(0.012 \mathrm{m}) \ln \left(\dfrac{0.018 \mathrm{m}}{ 145 \times 10^{-6} \mathrm{m}}\right) \\ &= \boxed{1157 \mathrm{V}} \end{align}
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