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, $\frac{E_{2}}{E_{1}}=\frac{1}{3^{2}}=\frac{1}{9}$.