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: 31

Answer

$E= -kQ\frac{4xa}{(x^2-a^2)^2}$.

Work Step by Step

Let the rightward direction be positive. The field due to +Q is positive, while the field due to -Q is negative. The negative charge is closer to point P so we expect the net field to point to the left. $$E=k\frac{Q}{(x+a)^2}+ k\frac{-Q}{(x-a)^2}$$ $$E=kQ(\frac{1}{(x+a)^2}-\frac{1}{(x-a)^2})$$ $$E= -kQ\frac{4xa}{(x^2-a^2)^2}$$ As expected, the field points to the left. The magnitude $E= \frac{4kQxa}{(x^2-a^2)^2}$.
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