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 469: 24

Answer

$1.4\times 10^8 N/C$.

Work Step by Step

The electric field due to the positive charge (#1) points away from it. The electric field due to the negative charge (#2) points toward it. Because both fields point in the same direction, the magnitudes are added. $$E=E_1+E_2$$ $$E=\frac{k|Q_1|}{r_1^2}+\frac{k|Q_2|}{r_2^2}$$ $$=(8.99\times10^9\frac{N\cdot m^2}{C^2})(\frac{8.0\times10^{-6}C}{(0.030m)^2}+\frac{5.8\times10^{-6}C}{(0.030m)^2})$$ $$=1.4\times 10^8 N/C$$ The direction is toward the negative charge.
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