Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 25 - The Electric Potential - Conceptual Questions: 1

Answer

(a) $\frac{U_{1}}{U_{2}}=6$ (b) $\frac{U_{1}}{U_{2}}=\frac{3}{2}$

Work Step by Step

The potential energy due to a charge placed a distance $r$ form the center of the positive point charge is $U=\frac{1}{4\pi \epsilon_{0}}$$\frac{Qq}{r}$ Potential energy of the charge in an electric field is $U=qEs$ Here, $s$ is the distance of charge from plate. (a) The potential energy due to a charge $q_{1}$placed a distance $r$ form the center of the positive point $Q$ charge is $U_{1}=\frac{1}{4\pi \epsilon_{0}}$$\frac{Qq_{1}}{r}$ The potential energy due to a charge $q_{2}$placed a distance $r$ form the center of the positive point $Q$ charge is $U_{2}=\frac{1}{4\pi \epsilon_{0}}$$\frac{Qq_{2}}{r}$ Substitute $q_{2}=\frac{q_{1}}{3}$ . $U_{2}=\frac{1}{4\pi \epsilon_{0}}$$\frac{Q\frac{q_{1}}{3}}{r}=\frac{1}{6}\frac{1}{4\pi \epsilon_{0}}$$\frac{Qq_{1}}{r}$ Calculate ratio $\frac{U_{1}}{U_{2}}$. $\frac{U_{1}}{U_{2}}=\frac{\frac{1}{4\pi \epsilon_{0}}\frac{Qq_{1}}{r}}{\frac{1}{6}\frac{1}{4\pi \epsilon_{0}}\frac{Qq_{1}}{r}}$ Hence, $\frac{U_{1}}{U_{2}}=6$ (b) Potential energy when the charge $q_{1}$ is located at a distance $s$ from the negative plate of a parallel-plate capacitor is, $U_{1}=q_{1}Es$ Potential energy when the charge $q_{2}=\frac{q_{1}}{3}$ is located at a distance $2s$ from the negative plate of a parallel-plate capacitor is, $U_{2}=\frac{q_{1}}{3}E(2s)=\frac{2}{3}q_{1}Es$ Calculate ratio $\frac{U_{1}}{U_{2}}$. $\frac{U_{1}}{U_{2}}=\frac{q_{1}Es}{\frac{2}{3}q_{1}Es}$ Hence, $\frac{U_{1}}{U_{2}}=\frac{3}{2}$.
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