Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 25 - Electric Charges and Forces - Exercises and Problems - Page 746: 38

Answer

$1.74\times 10^{-4}\;\rm N$,$308^\circ$

Work Step by Step

First of all, we need to draw the force diagram exerted on the 5 nC-charge. Recalling that the like charges repel and the unlike charge attract. As seen below, we put the 5 nC-charge at the origin and we numbered the three charges from 1 to 3. The net electric force exerted on the charge $q_1=\rm5\;nC$ in the $x$-direction is given by $$\sum F_x=F_{x,2\rm on1}-F_{x,3\rm on1}$$ We will neglect the signs of the charges since we are focusing on the charge's force direction. $$\sum F_x=\dfrac{kq_1q_2}{r^2_{2\rightarrow1}}\cos(90^\circ)+\dfrac{kq_1q_3}{r^2_{3\rightarrow1}}\cos\theta$$ where $\cos\theta=\dfrac{x_3}{r_{3\rightarrow1}}$ and $r=\sqrt{x_3^2+y_3^2}$, so that $$\sum F_x=0+\dfrac{kq_1q_3x_3}{(x_3^2+y_3^2)^\frac{3}{2}}$$ Plug the known; $$\sum F_x= \dfrac{(8.99\times 10^9)(5\times 10^{-9})(10\times 10^{-9})(0.03)}{(0.03^2+0.04^2)^\frac{3}{2}}$$ $$\sum F_x=\bf 1.08\times10^{-4}\;\rm N$$ The net electric force exerted on the charge $q_1=\rm 5\; nC$ in the $y$-direction is given by $$\sum F_y=-F_{y,2\rm on1}+F_{y,3\rm on1}$$ $$\sum F_y=-\dfrac{kq_1q_2}{r^2_{2\rightarrow1}}\sin(90^\circ)+\dfrac{kq_1q_3}{r^2_{3\rightarrow1}}\sin\theta$$ where $\sin\theta=\dfrac{y_3}{r_{3\rightarrow1}}$ and $r=\sqrt{x_3^2+y_3^2}$, so that $$\sum F_y=-\dfrac{kq_1q_2}{r^2_{2\rightarrow1}} +\dfrac{kq_1q_3y_3}{r^3_{3\rightarrow1}}$$ $$\sum F_y=-\dfrac{(8.99\times 10^9)(5\times 10^{-9})(10\times 10^{-9})}{0.04^2} +\dfrac{(8.99\times 10^9)(5\times 10^{-9})(10\times 10^{-9})(0.04)}{{(0.03^2+0.04^2)^\frac{3}{2}} }$$ $$\sum F_y=\bf -1.37\times 10^{-4}\;\rm N$$ Hence, the magnitude of the net force is given by $$\sum F_{\rm on 1}=\sqrt{(\sum F_x)^2+(\sum F_y)^2}$$ $$\sum F_{\rm on 1}=\sqrt{(1.08\times10^{-4})^2+(-1.37\times 10^{-4})^2}$$ $$\sum F_{\rm on 1}=\color{red}{\bf 1.74\times 10^{-4}}\;\rm N$$ And its direction is given by $$\tan\theta_{net}=\dfrac{\sum F_y}{\sum F_x}$$ $$\theta_{net}=\tan^{-1}\left[\dfrac{\sum F_y}{\sum F_x}\right]=\tan^{-1}\left[\dfrac{1.37\times 10^{-4}}{1.08\times 10^{-4}}\right]=\bf 51.8^\circ$$ where $\theta_{net}$ is in the fourth quadranet. And then its direction from $+x$-direction is $$\theta_{net}=360^\circ -51.8^\circ$$ $$\theta_{net}=\color{red}{\bf 308^\circ}\tag{CCW}$$ Counterclockwise from $+x$-direction.
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