Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 684: 18

Answer

$-2.21\times 10^{-5}C$

Work Step by Step

As given that there is a force of attraction between the charges and one point charge is positive so the required point charge is negative and its magnitude can be determined as follows: $Q=-\frac{r^2 F}{Kq}$ We plug in the known values to obtain: $Q=-\frac{(1.31)^2(0.975)}{8.99\times 10^9(8.44\times 10^{-6})}$ $Q=-2.21\times 10^{-5}C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.