Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 22 - Electric Fields - Problems - Page 655: 24c

Answer

$z = \frac{R}{\sqrt{2}}$

Work Step by Step

We can find the value of $z$ where the electric field is a maximum: $E = \frac{Qz}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}}$ $\frac{dE}{dz} = \frac{(Q)[4\pi~\epsilon_0~(z^2+R^2)^{3/2}]-(Qz)(3/2)[4\pi~\epsilon_0~(z^2+R^2)^{1/2}](2z)}{4\pi~\epsilon_0~(z^2+R^2)^{3/2}} = 0$ $(Q)[4\pi~\epsilon_0~(z^2+R^2)^{3/2}] = (Qz)(3/2)[4\pi~\epsilon_0~(z^2+R^2)^{1/2}](2z)$ $z^2+R^2 = 3z^2$ $2z^2 = R^2$ $z = \frac{R}{\sqrt{2}}$
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