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 779: 69

Answer

$ -2.29 \;\rm nC/m$

Work Step by Step

From the given information, it is obvious that the wire exerts an attractive radial force on the proton that forces it to move around it. This force is given by $$\sum F_r=qE=ma_r$$ where $q$ is the charge of the proton, $E$ is the electric field from the wire, $m$ is the mass of the proton, and $a_r$ the radial acceleration which is given by $v^2/r$, Thus, $$qE=m\dfrac{v^2}{r}$$ $$E=\dfrac{mv^2}{rq}\tag 1$$ And since the proton is positively charged, then the wire must be negatively charged, and we know that the electric field exerted by a line charge at a distance $r$ is given by $$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{2\lambda}{r}$$ Hence, the linear charge density of the wire is $$\lambda=\dfrac{(4\pi \epsilon_0)rE }{2}$$ Plug from (1); $$\lambda=\dfrac{(4\pi \epsilon_0)r }{2}\dfrac{mv^2}{rq}$$ $$\lambda=\dfrac{(4\pi \epsilon_0)mv^2}{ 2q}\tag2 $$ Now we need to find $v$, $$v=\dfrac{2\pi r}{T}=2\pi r f$$ where $f$ is the frequency. Plug into (2) $$\lambda=\dfrac{(4\pi \epsilon_0)m(2\pi rf)^2}{ 2q} $$ $$\lambda=\dfrac{(4\pi \epsilon_0)m (4\pi^2) r^2f^2}{ 2q} $$ Plug the known; $$\lambda=\dfrac{ (1.67\times 10^{-27}) (4\pi^2)(0.01)^2 (1\times 10^6)^2}{(9\times 10^9)( 2)(1.6\times 10^{-19})} $$ $$\lambda=\color{red}{\bf -2.29\times 10^{-9}}\;\rm C/m$$ The negative sign is due to the negative charge of the wire.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.