University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 23 - Electric Potential - Problems - Exercises - Page 778: 23.17


(a) $V_a$ = 0 (b) $V_b = - 171\times 10^{3} \,\text{V}$ (c) $W = -0.855\mathrm{J} $

Work Step by Step

(a) The potential at point $a$ will be the potential between the two charges \begin{align} V_{a} &= \dfrac{1}{4\pi \epsilon_o} \sum \frac{q}{r_{i}}\\ &= \dfrac{1}{4\pi \epsilon_o} \left(\frac{q_{1}}{r}+\frac{q_{2}}{r}\right)\\ & = \dfrac{1}{4\pi \epsilon_o} \left(\frac{+2.00 \mu \mathrm{C}}{r}+\frac{-2.00 \mu \mathrm{C}}{r}\right) \\ & = 0 \end{align} (b) From Physagros theorem, we could get the distance between $q_1$ and point $b$ by $$r_{1}=\sqrt{(.03)^{2}+(.03)^{2}}=0.042 \mathrm{m}$$ Now we could get the potential at point $b$ by \begin{align} V_{b} &= \dfrac{1}{4\pi \epsilon_o} \sum \frac{q}{r_{i}}\\ &= \dfrac{1}{4\pi \epsilon_o} \left(\frac{q_{1}}{r_1}+\frac{q_{2}}{r_2}\right)\\ & = \dfrac{1}{4\pi \epsilon_o} \left(\frac{+2.00 \mu \mathrm{C}}{0.042 \mathrm{m}}+\frac{-2.00 \mu \mathrm{C}}{0.03 \mathrm{m}}\right) \\ & =\boxed{ - 171\times 10^{3} \,\text{V}} \end{align} (c) The work done $W$ is given by \begin{aligned} W &=q_{3} \Delta V\\ & = q_3 \left(V_{a}-V_{b}\right) \\ &=\left(-5 \times 10^{-6}\right)\left(0 - [-171 \times 10^{3}]\right)\\ &=\boxed{-0.855 \mathrm{J} } \end{aligned}
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