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: 40

Answer

See the detailed answer below.

Work Step by Step

$$\color{blue}{\bf [a]}$$ We know that the electric field due to a continuous charge distribution is given by $$E=\dfrac{1}{4\pi \epsilon_0}\int_0^r\dfrac{1}{r^2}dq$$ Let's assume that the linear charge density of this rod is $\lambda$ which is given by $$\lambda=\dfrac{Q}{L}=\dfrac{dq}{dx}$$ where $dL$ is the length of the small black part in the figure below and $dx$ is its length. Thus, $$dq=\dfrac{Qdx}{L}\tag 1$$ So the electric field due to this small segment, as seen in the figure below, is given by $$dE=\dfrac{1}{4\pi \epsilon_0} \dfrac{dq}{(r-x)^2}$$ For the electric field from the rod, we need to take the integral for both sides, $$E_x=\dfrac{1}{4\pi \epsilon_0}\int_{-L/2}^{L/2} \dfrac{dq}{(r-x)^2}$$ Plug $dx$ from (1), $$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\int_{-L/2}^{L/2} \dfrac{1}{(r-x)^2}dx$$ $$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[ \dfrac{1}{(r-x) }\bigg]_{-L/2}^{L/2}$$ $$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[\dfrac{1}{r-(L/2)}-\dfrac{1}{r+(L/2)}\bigg] $$ $$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[\dfrac{r+(L/2)-[r-(L/2)]}{r^2-(L/2)^2} \bigg] $$ $$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[\dfrac{ L}{r^2-(L/2)^2} \bigg] $$ $$E_x=\dfrac{Q }{4\pi \epsilon_0}\bigg[\dfrac{1}{r^2-(L/2)^2} \bigg] $$ $$\boxed{E_x=\dfrac{1}{4\pi \epsilon_0}\left[\dfrac{Q}{r^2-\frac{L^2}{4}} \right] }$$ $$\color{blue}{\bf [b]}$$ when $r\gt \gt L$, then $r^2-\dfrac{L^2}{4}\approx r^2$. Thus, $$E_x=\dfrac{1 }{4\pi \epsilon_0}\bigg[\dfrac{Q}{r^2 } \bigg] $$ which is acceptable since at a long distance from the rod, it appears as a point charge. $$\color{blue}{\bf [c]}$$ Plug the known into the boxed formula above, $$E_x=\dfrac{(9.0\times 10^9) (3\times 10^{-9})}{0.03^2-0.025^2}$$ $$E_x=\color{red}{\bf 9.82\times 10^4}\;\rm N/C$$
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