Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 26 - The Electric Field - Exercises and Problems - Page 777: 37

Answer

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Work Step by Step

We know that the electric field exerted by an infinite line charge at a distance $r$ from it is given by $$E_{line}=\dfrac{1}{4\pi \epsilon_0}\;\dfrac{2|\lambda|}{r}$$ Now we need to draw the electric field direction as seen below. From the geometry of this graph, we can see that $\bullet$ $\theta_L=\theta_R=\theta$ $\bullet$ $r_L=r_R=\sqrt{(d/2)^2+y^2}$ $\bullet$ $\sin\theta=\dfrac{d/2}{r}=\dfrac{d}{ 2\sqrt{(d/2)^2+y^2}}$ $\bullet$ $\cos\theta=\dfrac{y}{r}=\dfrac{y}{\sqrt{(d/2)^2+y^2}}$ It is obvious, from the figure below, that the $y$ and $z$-components of the electric fields from both lines cancel each other and that the net electric field is in the $x$-direction. Hence, $$E_{net}=E_L+E_R$$ $$E_{net}= \dfrac{2 \lambda }{4\pi \epsilon_0 r}(-\sin\theta\;\hat i+\cos\theta\;\hat j)+\dfrac{2 \lambda }{4\pi \epsilon_0 r}(-\sin\theta\;\hat i-\cos\theta\;\hat j)$$ $$E_{net}= \dfrac{-4 \lambda }{4\pi \epsilon_0 r} \sin\theta\;\hat i$$ plug $\sin\theta$ from above, $$E_{net}= \dfrac{2 \lambda d}{4\pi \epsilon_0 r^2} \;\hat i $$ Hence, $$E_{net}= \dfrac{2 \lambda d}{4\pi \epsilon_0[ (d/2)^2+y^2]} \;\hat i$$ $$\boxed{E_{net}= \dfrac{8\lambda d}{4\pi \epsilon_0(d^2+4y^2)} \;\hat i}$$
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