Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 17 - Electric Potential - Problems - Page 497: 33

Answer

a. $6.6\times10^{-3}V$ b. $4.6\times10^{-3}V$ c. $-4.6\times10^{-3}V$

Work Step by Step

The electric potential due to the dipole is given by equation 17–6b. a. $V=\frac{kpcos \theta}{r^2}$ Referring to Figure 17-12, in this position the angle $\theta$ is $0^{\circ}$. $$V=\frac{kpcos \theta}{r^2}$$ $$=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(4.2\times10^{-30}C \cdot m)(cos0^{\circ})}{(2.4\times10^{-9}m)^2}$$ $$=6.6\times10^{-3}V$$ b. $V=\frac{kpcos \theta}{r^2}$ Referring to Figure 17-12, in this position the angle $\theta$ is $45^{\circ}$. $$V=\frac{kpcos \theta}{r^2}$$ $$=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(4.2\times10^{-30}C \cdot m)(cos45^{\circ})}{(2.4\times10^{-9}m)^2}$$ $$=4.6\times10^{-3}V$$ c. $V=\frac{kpcos \theta}{r^2}$ Referring to Figure 17-12, in this position the angle $\theta$ is $135^{\circ}$. $$V=\frac{kpcos \theta}{r^2}$$ $$=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(4.2\times10^{-30}C \cdot m)(cos135^{\circ})}{(2.4\times10^{-9}m)^2}$$ $$=-4.6\times10^{-3}V$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.