Answer
$2.0\times10^{5}N/C$
Work Step by Step
E=$\frac{F}{q}$(F is the magnitude of force and q is the charge of the electron)
$= \frac{3.2\times10^{-14}N}{1.602\times10^{-19}C}\approx 2.0\times10^{5}N/C.^{\circ}$
College Physics (7th Edition)
by Wilson, Jerry D.; Buffa, Anthony J.; Lou, Bo
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: 22
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