Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 22 - Electric Charges and Forces - Exercises and Problems - Page 626: 59

Answer

$q = 33.2~nC$

Work Step by Step

We can find the spring constant $K$: $Kx = mg$ $K = \frac{mg}{x}$ $K = \frac{(0.0010~kg)(9.8~m/s^2)}{0.010~m}$ $K = 0.98~N/m$ We can find the magnitude of the charge on each bead: $F = \frac{kq^2}{r^2} = Kx$ $q^2 = \frac{Kx~r^2}{k}$ $q = \sqrt{\frac{Kx~r^2}{k}}$ $q = \sqrt{\frac{(0.98~N/m)(0.0050~m)~(0.045~m)^2}{9.0\times 10^9~N~m^2/C^2}}$ $q = 33.2\times 10^{-9}~C$ $q = 33.2~nC$
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