Answer
$q_1 = -20~nC$
Work Step by Step
We can write Coulomb's Law:
$F = \frac{k~q_1~q_2}{r^2}$
Since $F \propto \frac{1}{r^2},$ then $\vert q_1 \vert = (4)(5.0~nC) = 20~nC$
Therefore, $~~q_1 = -20~nC$
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
by Knight, Randall D.
Published by
Pearson
ISBN 10:
0133942651
ISBN 13:
978-0-13394-265-1
Chapter 22 - Electric Charges and Forces - Exercises and Problems - Page 626: 49
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