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 28 - The Electric Potential - Conceptual Questions - Page 832: 12

Answer

See the figure below.

Work Step by Step

The net electric potential at some point at the same line between two charges is given by $$V=V_1+V_2=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$$ when that line is $x$-axis, $$V=\dfrac{kq_1}{x_1}+\dfrac{kq_2}{x_2}$$ We chose $q_1=+2q$ to be at the origin at which $x=0$, and hence the charge $q_2=-q$ is at $x=3$. $$V=\dfrac{2k q}{x_1}+\dfrac{-kq}{x_2}$$ We have here 3 possibilities where we can get a net electric potential energy of zero; A point to the left from $q_1$, a point between $q_1$ and $q_2$, and a point to the right from $q_2$. $\Rightarrow$ When the point is to the left from $q_1$, $x_1=|x|$, and $x_2=|x+3|$ $$V=\dfrac{2k q}{|x|}+\dfrac{-kq}{|x+3|}=0$$ $$ \dfrac{2k q|x+3|-kq|x|}{|x+3|\cdot|x|}=0$$ Hence, $$2|x+3|-|x|=0$$ Thus, $$2x+6=\pm x$$ where the result is $x=-2$ or $x=-6$ and since we are dealing with distances, not position, this result is dismissed because there are no distances with negative. $\Rightarrow$ When the point is between the two charges, $x_1=|x|$, and $x_2=|3-x|$ $$V=\dfrac{2k q}{|x|}+\dfrac{-kq}{|3-x|}=0$$ $$ \dfrac{2 |3-x| -|x|}{|3-x|\cdot |x|}=0$$ Hence, $$2 |3-x| -|x|=0$$ $$ 6-2x =\pm x$$ Thus, $x=2$ or $x=6$ those are real roots. See the figure below, $\Rightarrow$ When the point is to the right from $q_2$, $x_1=|x|$, and $x_2=|x-3|$ $$V=\dfrac{2k q}{|x|}+\dfrac{-kq}{|x-3|}=0$$ $$ \dfrac{2 |x-3|- |x|}{|x-3|\cdot|x|}=0$$ Hence, $$2|x-3|-|x|=0$$ Thus, $$2x-6=\pm x$$ where the result is $x= 2$ or $x= 6$ those are real roots. From all the above, it is obvious that there are only two points at which the net electric potential energy is zero, at $x=2$ and at $x=6$, as shown in the figure below.
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