Answer
$E = 2.03\times 10^{-7}~N/C$
Work Step by Step
We can find the magnitude of the electric field:
$q~E = Mg$
$E = \frac{Mg}{q}$
$E = \frac{(6.64\times 10^{-27}~kg)(9.8~m/s^2)}{(2)(1.6\times 10^{-19}~C)}$
$E = 2.03\times 10^{-7}~N/C$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 22 - Electric Fields - Problems - Page 656: 44a
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