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: 5

Answer

(a) $E = 36~N/C$ (b) $E = 18~N/C$

Work Step by Step

(a) We can find the electric field strength at the point $(10~cm, 0)$ which is along the axis of the dipole. $E = \frac{q~d}{2\pi~\epsilon_0~r^3}$ $E = \frac{(1.0\times 10^{-9}~C)(0.0020~m)}{(2\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3}$ $E = 36~N/C$ (b) We can find the electric field strength at the point $(0,10 ~cm)$ which is along the y-axis. $E = \frac{q~d}{4\pi~\epsilon_0~r^3}$ $E = \frac{(1.0\times 10^{-9}~C)(0.0020~m)}{(4\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3}$ $E = 18~N/C$
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