Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.9 - The Divergence Theorem - 16.9 Exercise - Page 1146: 23


$div E=0$ for the electric field $E(x) =\dfrac{\epsilon Q x}{|x^3|}$

Work Step by Step

$E(x) =\dfrac{\epsilon Q x}{|x^3|}$ This implies that $E(x)=\dfrac{\epsilon Q (xi+yj+zk) }{(x^2+y^2+z^2)^{3/2}}= \epsilon Q F$ Here, we have $f(x,y,z)=\dfrac{1}{(x^2+y^2+z^2)^{3/2}}$ Also, $\nabla f=f_xi+f_y j+f_z k$ Thus, we have $\nabla f=-\dfrac{3x}{(x^2+y^2+z^2)^{5/2}}i-\dfrac{3y}{(x^2+y^2+z^2)^{5/2}} j-\dfrac{3z}{(x^2+y^2+z^2)^{5/2}}k$ or, $=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk)$ Therefore, $div F=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) \cdot (xi+yj+zk)+\dfrac{3}{(x^2+y^2+z^2)^{3/2}}=\dfrac{3}{(x^2+y^2+z^2)^{3/2}}-\dfrac{3}{(x^2+y^2+z^2)^{3/2}}=0$ This implies that $div E=0$ for the electric field $E(x) =\dfrac{\epsilon Q x}{|x^3|}$
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