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: 14

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ Since the charges are uniformly distributed in the disk, the net electric field at a point of $x$ that passes through its center (axis) must be in the $x$-direction. The $y$-components and $z$-components from the electric field are zeros. Hence, the net electric field in this situation is then given by $$E_{disk,x}=\dfrac{\eta}{2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] $$ where $\eta=\dfrac{Q}{A}=\dfrac{Q}{\pi r^2}$ $$E_{disk,x}=\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right] $$ Note that the electric field from a solid disk is the sum of the electric field from a group of rings and this formula is found by integrating the formula of rind; you can see that in your textbook. Now we have two positively charged disks, as seen below. So the net electric field at point P just in the middle between them must be zero. $$E_{net}= (E_1-E_2)\hat i$$ $$E_{net}= \left(\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right]-\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x}{\sqrt{x^2+r^2}}\right]\right)\hat i$$ Noting that $|Q_1|=|Q_2|=Q$, $r_1=r_2=r=5$ cm, and $x_1=x_2=x=10$ cm. Thus, $$E_{net}= \color{red}{\bf0}\;{\rm N/C} $$ $$\color{blue}{\bf [b]}$$ Now at the new point K, where $x_1=0.05$ m and $x_2=0.15$ m, $$E_{net}= \left(\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x_1}{\sqrt{x_1^2+r^2}}\right]-\dfrac{Q}{2\pi r^2\epsilon_0 }\left[1-\dfrac{x_2}{\sqrt{x_2^2+r^2}}\right]\right)\hat i$$ $$E_{net}=\dfrac{Q}{2\pi r^2\epsilon_0 } \left( 1-\dfrac{x_1}{\sqrt{x_1^2+r^2}} - 1+\dfrac{x_2}{\sqrt{x_2^2+r^2}} \right)\hat i$$ $$E_{net}=\dfrac{Q}{2\pi r^2\epsilon_0 } \left( -\dfrac{x_1}{\sqrt{x_1^2+r^2}} +\dfrac{x_2}{\sqrt{x_2^2+r^2}} \right)\hat i$$ Plugging the known; $$E_{net}= \dfrac{ (50\times 10^{-9})}{2 \pi (0.05)^2(8.85\times 10^{-12})}\left(-\dfrac{(0.05)}{\sqrt{(0.05)^2+(0.05)^2}}+\dfrac{(0.15)}{\sqrt{(0.15)^2+(0.05)^2}}\right) \hat i$$ $$E_{net}=(\color{red}{\bf 8.7\times 10^4}\;{\rm N/C})\hat i\tag{Rightward}$$
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