College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 16 - Problems - Page 615: 42


$E = 1.00\times 10^6~N/C$

Work Step by Step

We can find the distance $r$ from each charge to point B: $r = \sqrt{(0.300~m)^2+(0.150~m)^2}$ $r = 0.3354~m$ Let $\theta$ be the angle above the horizontal of the line from each point charge to point B. By symmetry, the horizontal component of the electric field due to each point charge cancels out. The net electric field at point B is the sum of the vertical components of the electric field due to each point charge: $E = 2\times \frac{kq}{r^2}~sin~\theta$ $E = 2\times \frac{(9.0\times 10^9~N~m^2/C^2)(7.00\times 10^{-6}~C)}{(0.3354~m)^2}~(\frac{0.300}{0.3354})$ $E = 1.00\times 10^6~N/C$
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