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, $\frac{U_{1}}{U_{2}}=\frac{3}{2}$.
