Answer
$$3.6\times 10^{-6}N$$
Work Step by Step
$F_{BC}=\frac{k\times q_{B}\times q_{C}}{d^{2}}$
$=\frac{9\times10^{9}\times4\times10^{-9}\times4\times10^{-9}}{(2\times10^{-1})^{2}}$
$=\frac{72\times10^{-9}}{10^{-2}}$
$=3.6\times10^{-6}N$
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 21 - Coulomb's Law - Problems - Page 628: 48c
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