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 26 - The Electric Field - Exercises and Problems - Page 776: 31

Answer

See the detailed answer below.

Work Step by Step

First, we need to draw the electric field direction at the black dot, as seen below. Thus, from the figure below, we can see that $$E_1=-\dfrac{k|q_1|}{r_1^2}\;\hat j$$ Plug the known; $$E_1=-\dfrac{kQ}{L^2}\;\hat j \tag 1$$ $$E_2=\dfrac{k|q_2|}{r_2^2}(\cos\theta\;\hat i+\sin\theta\;\hat j)$$ where $\sin\theta= L/r_2$, $\cos\theta=L/r_2$. Thus, $$E_2=\dfrac{k|q_2|}{r_2^3}(L\;\hat i+L\;\hat j)$$ where $r_2=\sqrt{L^2+L^2}=\sqrt{2}L$ $$E_2=\dfrac{4kQL}{(\sqrt{2}L)^3}(1\;\hat i+1\;\hat j)$$ $$E_2=\dfrac{4kQ }{2^{3/2}{L^2}} \;\hat i+\dfrac{4kQ }{2^{3/2}{L^2}}\;\hat j \tag 2$$ $$E_3= -\dfrac{kq_3}{r_3^2}\;\hat i$$ Plug the known; $$E_3= -\dfrac{kQ}{L^2}\;\hat i\tag 3$$ Therefore, the net electric field in a component form is given by $$E_{net}=E_1+E_2+E_3$$ Plugging from above (1), (2), (3), $$E_{net}= \left[-\dfrac{kQ}{L^2}\;\hat j\right]+\left[\dfrac{4kQ }{2^{3/2}{L^2}} \;\hat i+\dfrac{4kQ }{2^{3/2}{L^2}}\;\hat j \right]+\left[-\dfrac{kQ}{L^2}\;\hat i\right]$$ $$E_{net}= \left[\dfrac{4kQ }{2^{3/2}{L^2}} -\dfrac{kQ}{L^2} \right] \;\hat i+\left[ \dfrac{4kQ }{2^{3/2}{L^2}}-\dfrac{kQ}{L^2}\;\right]\hat j$$ $$E_{net}= \left[\dfrac{4kQ -2^{3/2}kQ}{2^{3/2}{L^2}} \right] \;\hat i+\left[ \dfrac{4kQ -2^{3/2}kQ}{2^{3/2}{L^2}} \;\right]\hat j$$ $$E_{net}= \left[\dfrac{kQ( 4-2\sqrt2)}{2\sqrt{2}{L^2}} \right] \;\hat i+\left[ \dfrac{kQ( 4-2\sqrt2)}{2\sqrt{2}{L^2}} \;\right]\hat j$$ $$\boxed{E_{net}= \left[\dfrac{kQ( \sqrt{2}-1)}{{L^2}} \right] \;\hat i+\left[ \dfrac{kQ( \sqrt{2}-1)}{{L^2}} \;\right]\hat j}$$
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