Introduction to Electrodynamics 4e

Published by Pearson Education
ISBN 10: 9332550441
ISBN 13: 978-9-33255-044-5

Chapter 2 - Section 1.3 - The Electric Field - Problem - Page 62: 2


\begin{align*}\text{Electric field, E}=\frac{1}{4\,\pi\,\epsilon_{0}}\,\frac{qd}{[z^2+(d/2)]^{3/2}}\,\hat{x}\end{align*}

Work Step by Step

Here, $E_1$ is the field due to the positive charge alone, and $E_2$ is the field due to the negative charge alone (in figure). The vertical components of the field cancel each other out, leaving a net horizontal component along the positive x-axis - \begin{align*} E_x = 2\,\frac{1}{4\,\pi\,\epsilon_{0}}\,\frac{q}{r^2}\,\sin\theta\tag{1} \end{align*} where the symbols are as described in the figure. Also, from figure $r=\sqrt (z^2+(d/2)^2)$ and $\sin\theta=d/2r$. Thus, substituting in equation (1) the electric field is \begin{align*} \text{E}=\frac{1}{4\,\pi\,\epsilon_{0}}\,\frac{qd}{[z^2+(d/2)]^{3/2}}\,\hat{x}\end{align*}
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