Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 16 - Electric Charge and Electric Field - Problems - Page 470: 32

Answer

65 cm.

Work Step by Step

The net field is zero at point P. Set the magnitudes of the fields created by Q1 and by Q2 to be equal. Let x be a positive number representing the desired distance. $$k\frac{|Q1|}{x^2}= k\frac{|Q2|}{(x+0.12m)^2}$$ $$k\frac{32\times10^{-6}C}{x^2}= k\frac{45\times10^{-6}C }{(x+0.12m)^2}$$ The constant k cancels out. Solve for x. $$x=(0.12m)\frac{\sqrt{32\times10^{-6}C}}{\sqrt{45\times10^{-6}C}-\sqrt{32\times10^{-6}C}}$$ $$x=64.57\times10^{-2}m\approx 65\;cm$$
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