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: 44

Answer

See the detailed answer below.

Work Step by Step

$$\color{blue}{\bf [a]}$$ First, we need to sketch the problem as shown below. The electric field exerted by the small black segment of the rod, which has a charge of $dq$ and a length of $ds$, at the center of the circle (which we chose as our origin), is $dE$. $$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{dq}{R^2}$$ We can also observe that the $y$-components of the small segments like ours cancel each other out, while the $x$-components add up in the positive (x)-direction. So the electric field exerted by the small black segment in the $x$-direction is given by $$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{dq}{R^2}\cos\theta\;\hat i\tag 1$$ Let's assume that the linear charge density of the rod is $\lambda$, so $$\lambda=\dfrac{Q}{L}=\dfrac{dq}{ds}$$ where $ds$ is the length of the small arc (black segment in the figure below) and we know that the arc length is given by $R\theta$ where $\theta$ in radians, so $$\lambda=\dfrac{Q}{L}=\dfrac{dq}{Rd\theta}$$ Hence, $$dq=\dfrac{QRd\theta}{L}$$ Plug into (1), $$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{QR}{LR^2}d\theta\cos\theta\;\hat i $$ $$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q }{LR }\cos\theta d\theta\;\hat i $$ Now to find the net electric field of the whole semi-circle rod, we need to integrate both sides relative to $\theta$ where $\theta$ varies from $-\pi/2$ to $\pi/2$. $$E_x=\int dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q }{LR }\int_{-\pi/2}^{\pi/2}\cos\theta d\theta $$ $$E_x= \dfrac{1}{4\pi \epsilon_0}\dfrac{Q }{LR }\left[ \sin\theta\right]_{-\pi/2}^{\pi/2} $$ $$E_x= \dfrac{1}{4\pi \epsilon_0}\dfrac{Q }{LR }\left[ \sin(\pi/2)-\sin(-\pi/2)\right] $$ $$ E_x= \dfrac{2Q}{4\pi \epsilon_0 LR} $$ Now we need to find the radius $R$ in terms of $L$, where we know that the length of a semi-circle is given by $L=\pi R$, so $R=L/\pi$ $$ \boxed{E = \dfrac{2\pi Q}{(4\pi \epsilon_0) L^2} \;\hat i } $$ $$\color{blue}{\bf [b]}$$ Plugging the known into the boxed formula above, $$ E = \dfrac{2\pi (9\times 10^9)(30\times 10^{-9})}{ (0.10)^2} \;\hat i $$ $$ E =(\color{red}{\bf1.7\times 10^5}\;{\rm N/C})\;\hat i$$
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