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 27 - Gauss's Law - Exercises and Problems - Page 808: 53

Answer

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Work Step by Step

The orange dashed circle in the figure below are the 3 Gaussian spheres. According to Gauss's law, $$\oint \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$ According to the uniform charge distribution in the shell, the shell has spherical symmetry. So the electric field is uniform. $$EA=\dfrac{Q_{in}}{\epsilon_0}$$ Hence, $$E=\dfrac{Q_{in}}{\epsilon_0A}$$ $$\color{blue}{\bf [a]}$$ when $r\gt R_{\rm out}$; $$E=\dfrac{Q_{in}}{\epsilon_0A}=\dfrac{Q}{4\pi r^2\epsilon_0}$$ Hence, $$\boxed{\vec E=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{r^2}\;\hat r}\tag {Outward}$$ $$\color{blue}{\bf [b]}$$ when $r\lt R_{\rm in}$; $$E=\dfrac{Q_{in}}{\epsilon_0A}=\dfrac{0}{4\pi r^2\epsilon_0}$$ Hence, $$\boxed{\vec E=\bf 0\;\rm N/C}$$ $$\color{blue}{\bf [c]}$$ when $R_{\rm in}\lt r\lt R_{\rm out}$; $$E=\dfrac{Q_{in}}{\epsilon_0A}$$ $$E=\dfrac{Q_{in}}{4\pi r^2\epsilon_0}\tag 1$$ where the volume charge density of the shell is given by $$\rho_s=\dfrac{Q}{V}=\dfrac{Q}{\frac{4}{3}\pi (R_{\rm out}^3 -R_{\rm in}^3 )}=\dfrac{Q_{in}}{\frac{4}{3}\pi (r^3-R_{\rm in}^3)}$$ where $V=V_{\rm out}-V_{\rm in}$ Hence, $$Q_{in}=\dfrac{\frac{4}{3}\pi (r^3-R_{\rm in}^3)Q }{\frac{4}{3}\pi(R_{\rm out}^3 -R_{\rm in}^3 ) }=\dfrac{(r^3-R_{\rm in}^3) Q }{ (R_{\rm out}^3 -R_{\rm in}^3 ) }$$ Plug into (1), $$E=\dfrac{1}{4\pi r^2\epsilon_0}\dfrac{(r^3-R_{\rm in}^3) Q}{ (R_{\rm out}^3 -R_{\rm in}^3 ) }$$ $$\boxed{\vec E=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q(r^3-R_{\rm in}^3)}{ r^2(R_{\rm out}^3 -R_{\rm in}^3 )}\;\hat r}\tag {Outward}$$ $$\color{blue}{\bf [d]}$$ when $r=R_{\rm in}$, $$ E=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q(R_{\rm in}^3-R_{\rm in}^3)}{ r^2(R_{\rm out}^3 -R_{\rm in}^3 )} =\bf 0\;\rm N/C$$ which is the same result of (b), And when $r=R_{\rm out}$, $$ E=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q(R_{\rm out}^3-R_{\rm in}^3)}{ r^2(R_{\rm out}^3 -R_{\rm in}^3 )} =\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{r^2}$$ which is the same result of (a), So the previous boxed formula, in part (c), matches at both the inner and outer boundaries.
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