Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 30 - Current and Resistance - Exercises and Problems - Page 889: 51

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ Since the current is radially outward from the inner surface to the outer surface, the current density at a distance $r$ (where $a\lt r\lt b$) from the center of the sphere is given by $$J =\dfrac{I}{A}$$ where $J$ is also given by $\sigma E$ Hence, $$\sigma E =\dfrac{I}{A}$$ Thus the electric field is given by $$ E =\dfrac{I}{\sigma A}$$ And if we assume that there is a spherical surface inside our hollow sphere, then $A=4\pi r^2$ $$ \boxed{E =\dfrac{I}{4\pi \sigma r^2}}$$ $$\color{blue}{\bf [b]}$$ At the inner surface, when $r=a$ $$ E_{\rm inner} =\dfrac{I}{4\pi \sigma_{\rm copper} a^2} $$ Plug the known; $$ E_{\rm inner} =\dfrac{25}{4\pi (6\times 10^7)(0.01)^2} $$ $$ E_{\rm inner} =\color{red}{\bf 3.3\times 10^{-4}}\;\rm V/m$$ At the outer surface, when $r=b$ $$ E_{\rm outer } =\dfrac{I}{4\pi \sigma_{\rm copper} b^2} $$ Plug the known; $$ E_{\rm outer } =\dfrac{25}{4\pi (6\times 10^7)(0.025)^2} $$ $$ E_{\rm outer } =\color{red}{\bf 5.3\times 10^{-5}}\;\rm V/m$$
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