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 775: 9

Answer

See the detailed answer below.

Work Step by Step

First, we need to draw these 3 dots and the electric field direction. We know that the electric field direction is away from a positive charge and is toward the negative charge, see the second figure below. We know that the electric field exerted on a point at distance $r$ from the center of a long charged rod is given by $$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+\left(\frac{L}{2}\right)^2}}$$ We have two glass rods, we name one of them, the upper, rod A, and the other rod B. Thus, at point 1, the net electric field is given by $$E_1=E_{A}+E_{B}$$ We chose upward to be our positive direction, $$E_1=-\dfrac{1}{4\pi \epsilon_0} \dfrac{|Q_{A}|}{r_{A}\sqrt{r_{A}^2+ \left(\frac{L_{A}}{2}\right)^2}}\;\hat j+\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q_{B}|}{r_{B}\sqrt{r_{B}^2+\left(\frac{L_{B}}{2}\right)^2}}\;\hat j $$ Recall that $|Q_{A}|=|Q_{B}|=Q$, and $L_{A}=L_{B}=L$ $$E_1=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{A}\sqrt{r_{A}^2+\left(\frac{L}{2}\right)^2}}\;-\dfrac{1}{r_{B}\sqrt{r_{B}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j\tag 1$$ Noting that $r_{A}=r_{A\rightarrow P_1}=0.01$ m while $r_{B}=r_{B\rightarrow P_1}=0.03$ m, $$E_1=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.01)\sqrt{(0.01)^2+\left(\frac{0.1}{2}\right)^2}}-\dfrac{1}{(0.03)\sqrt{(0.03)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$ $$E_1=(\color{red}{\bf -1.25\times 10^5}\;\hat j)\;\rm N/C$$ By the same approach, using (1), at point 2, the net electric field is given by $$E_2=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{A}\sqrt{r_{A}^2+\left(\frac{L}{2}\right)^2}}-\dfrac{1}{r_{B}\sqrt{r_{B}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j $$ Noting that $r_{A}=r_{A\rightarrow P_2}=0.02$ m while $r_{B}=r_{B\rightarrow P_2}=0.02$ m, $$E_2=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}-\dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$ $$E_2=(\color{red}{\bf 0}\;\hat j)\;\rm N/C$$ By the same approach, using (1), at point 3, the net electric field is given by $$E_3=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{A}\sqrt{r_{A}^2+\left(\frac{L}{2}\right)^2}}\;+\dfrac{1}{r_{B}\sqrt{r_{B}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j $$ Noting that $r_{A}=r_{A\rightarrow P_3}=0.03$ m while $r_{B}=r_{B\rightarrow P_3}=0.01$ m, $$E_3=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.03)\sqrt{(0.03)^2+\left(\frac{0.1}{2}\right)^2}}-\dfrac{1}{(0.01)\sqrt{(0.01)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$ $$E_3=(\color{red}{\bf 1.25\times 10^5}\;\hat j)\;\rm N/C$$
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