College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 15 - Electric Charge, Forces, and Fields - Learning Path Questions and Exercises - Exercises - Page 557: 21


(a) If the distance from the charge is doubled the magnitude of electric field (2) decreased to the initial value (b) magnitude of new electric field at twice the distance will be $E_{2r}=2.5\times10^{-15}N/C$

Work Step by Step

(a) From the definition of the electric field magnitude of the electric field due to a point charge is $E= \frac{kq}{r^2}$, so electric field is inversely proportional to square of distance from the charge. So if distance increases electric field decreases. (b) given that magnitude of electric field at distance $ r$ is $E_{r}= \frac{kq}{r^2}= 1.0\times10^{-14}N/C$ now if the distance is doubled electric field will be $E_{2r}= \frac{kq}{(2r)^2}=\frac{1}{4}\times \frac{kq}{r^2}$ $E_{2r}=\frac{1}{4}\times1.0\times10^{-14}N/C=0.25\times10^{-14}N/C$ $E_{2r}=2.5\times10^{-15}N/C$
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