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: 6

Answer

(a) $q = 2.0~nC$ (b) $E = 180~N/C$

Work Step by Step

(a) The point $(0~cm, 10~cm)$ is along the axis of the dipole. We can find the charge $q$. $E = \frac{q~d}{2\pi~\epsilon_0~r^3}$ $q = \frac{2\pi~\epsilon_0~r^3~E}{d}$ $q = \frac{(2\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3(360~N/C)}{0.010~m}$ $q = 2.0~nC$ (b) We can find the electric field strength at the point $(10 ~cm, 0~cm)$ which is along the x-axis. $E = \frac{q~d}{4\pi~\epsilon_0~r^3}$ $E = \frac{(2.0\times 10^{-9}~C)(0.0010~m)}{(4\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3}$ $E = 180~N/C$
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