Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 23 - The Electric Field - Exercises and Problems: 12

Answer

$Q = 0.056~nC$

Work Step by Step

We can use the equation for the electric field strength at a distance $r$ from a charged wire to find the linear charge density: $E = \frac{\lambda}{2\pi ~\epsilon_0~r}$ $\lambda = E~2\pi ~\epsilon_0~r$ $\lambda = (2000~N/C)~(2\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.050~m)$ $\lambda = 5.6\times 10^{-9}~C/m$ We can find the charge on a 1.0 cm segment of wire. $\lambda = \frac{Q}{L}$ $Q = \lambda~L$ $Q = (5.6\times 10^{-9}~C/m)(0.010~m)$ $Q = 0.056\times 10^{-9}~C$ $Q = 0.056~nC$
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.