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: 45

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 $dE$. $$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{dq}{R^2}(\cos\theta\;\hat i+\sin\theta\;\hat j) \tag 1$$ 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), $$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{QRd\theta}{LR^2}(\cos\theta\;\hat i+\sin\theta\;\hat j) $$ $$dE=\dfrac{Q}{(4\pi \epsilon_0)LR }(\cos\theta d\theta\;\hat i+\sin\theta d\theta\;\hat j) $$ Now we need to find the radius $R$ in terms of $L$, where we know that the length of a quarter-circle is given by $L=\pi R/2$, so $R=2L/\pi$ $$\boxed{dE=\dfrac{\pi Q}{(4\pi \epsilon_0)2L^2 }(\cos\theta d\theta\;\hat i+\sin\theta d\theta\;\hat j) }$$ And in terms of $R$, $$\boxed{dE=\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }(\cos\theta d\theta\;\hat i+\sin\theta d\theta\;\hat j) }$$ $$\color{blue}{\bf [b]}$$ 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 $0$ to $\pi/2$. $$E=\int dE=\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }\left[\int_0^{\pi/2} \cos\theta d\theta\;\hat i+\int_0^{\pi/2}\sin\theta d\theta\;\hat j\right] $$ $$\color{blue}{\bf [c]}$$ $$E =\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }\left[(\sin(\pi/2)-\sin 0)\;\hat i- (\cos(\pi/2)-\cos0)\;\hat j\right] $$ $$\boxed{\vec E =\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }\left[\hat i+\hat j\right] }$$
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