College Physics (4th Edition)

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

Chapter 16 - Problems - Page 617: 72


(a) $\Phi = E~(4\pi~r^2)$ (b) $c = \frac{1}{\epsilon_0}$

Work Step by Step

(a) We can find an expression for the electric flux: $\Phi = E \cdot A = E~(4\pi~r^2)$ (b) Note that $~E = \frac{q}{4\pi~\epsilon_0~r^2}$ We can find the constant of proportionality $c$: $\Phi = c~q$ $E~(4\pi~r^2) = c~q$ $\frac{4\pi~r^2~q}{4\pi~\epsilon_0~r^2} = c~q$ $\frac{q}{\epsilon_0} = c~q$ $c = \frac{1}{\epsilon_0}$
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