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

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