College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 16 - Problems - Page 618: 93


$q = -1.5~nC$

Work Step by Step

Let $q_1$ be the charge at the origin. Then $r_1 = 1.0~m$ and $r_2 = 0.50~m$. Note that $r_1 = 2~r_2$ At the point $x = 1.0~m$, the sum of the electric fields due to the two point charges is zero. We can find $q$: $E = \frac{k~q_1}{r_1^2} + \frac{k~q}{r_2^2} = 0$ $\frac{k~q}{r_2^2} = -\frac{k~q_1}{r_1^2}$ $\frac{q}{r_2^2} = -\frac{q_1}{r_1^2}$ $q = -\frac{q_1~r_2^2}{r_1^2}$ $q = -\frac{q_1~r_2^2}{(2~r_2)^2}$ $q = -\frac{q_1}{4}$ $q = -\frac{+6.0~nC}{4}$ $q = -1.5~nC$
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