College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 16 - Problems - Page 617: 73

Answer

(a) $E = \frac{q}{\epsilon_0~4\pi~r^2}$ (b) $E = 0$

Work Step by Step

(a) We can imagine a spherical shell with radius $r$ outside the spherically symmetrical charge distribution where this charge distribution has a total charge of $q$. We can use Gauss' law to find an expression for the electric field at the location of the spherical shell: $\Phi = E \cdot A = \frac{q}{\epsilon_0}$ $E~(4\pi~r^2) = \frac{q}{\epsilon_0}$ $E = \frac{q}{\epsilon_0~4\pi~r^2}$ Note that this is the same expression for the electric field if all the charge $q$ were concentrated into a point charge $q$. (b) We can imagine a spherical shell with radius $r$ inside the spherically symmetrical charge distribution where this charge distribution has a total charge of $q = 0$ inside the radius $r$. We can find use Gauss' law to find an expression for the electric field at the location of the spherical shell: $\Phi = E \cdot A = \frac{q}{\epsilon_0}$ $E~(4\pi~r^2) = \frac{q}{\epsilon_0}$ $E = \frac{q}{\epsilon_0~4\pi~r^2}$ $E = \frac{0}{\epsilon_0~4\pi~r^2}$ $E = 0$
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