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 - General Problems - Page 472: 63

Answer

x = 2.41d.

Work Step by Step

The electric field strength drops off quickly. Any place where the field is zero is closer to the negative charge, since it is weaker. In between the two charges, the fields due to the two charges both point to the right, and cannot cancel. At point P, the fields will cancel, so set the magnitudes equal. $$k\frac{Q}{(d+x)^2}= k\frac{Q/2}{x^2}$$ $$2x^2=(x+d)^2$$ $$x=(\sqrt{2}+1)d\approx 2.41d$$
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