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: 10

Answer

(a) $\frac{V_{2}}{V_{1}}=\frac{1}{3}$ (b) $\frac{E_{2}}{E_{1}}=\frac{1}{9}$

Work Step by Step

(a) $\frac{V_{2}}{V_{1}}=\frac{r_{1}}{r_{2}}$ (since, charge $q$ is same and $V=\frac{q}{4 \pi \epsilon_{0}r}$) Therefore, $\frac{V_{2}}{V_{1}}=\frac{1}{3}$. (a) $\frac{V_{2}}{V_{1}}=\frac{r_{1}}{r_{2}}$ (since, charge $q$ is same and $V=\frac{q}{4 \pi \epsilon_{0}r}$) Therefore, $\frac{V_{2}}{V_{1}}=\frac{1}{3}$. (b) $\frac{E_{2}}{E_{1}}=\frac{r_{1}^{2}}{r_{2}^{2}}$ (since, charge $q$ is same and $E=\frac{q}{4 \pi \epsilon_{0}r^{2}}$) Therefore, $\frac{E_{2}}{E_{1}}=\frac{1}{3^{2}}=\frac{1}{9}$.
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