A point charge of +6.0 nC is in air. (a) What is the radius of the equipotential sphere at 90 V? …

Step 1. The first radius. $kq = 9.0 \times 10^{9}\ \text{N m}^2/\text{C}^2 \times 6.0 \times 10^{-9}\ \text{C} = 54\ \text{V m}$. From $V = kq/r$ we want $r$, so $r = \dfrac{kq}{V} = \dfrac{54\ \text{V m}}{90\ \text{V}} = 0.60\ \text{m}$.

Step 2. The second radius. $r = \dfrac{54\ \text{V m}}{45\ \text{V}} = 1.20\ \text{m}$.

Step 3. The fields. $E = \dfrac{kq}{r^2} = \dfrac{54\ \text{V m}}{(0.60\ \text{m})^2} = 150\ \text{V/m}$. At the farther sphere, $E = \dfrac{54\ \text{V m}}{(1.20\ \text{m})^2} = 37.5\ \text{V/m}$.

Step 4. The ratio. $\dfrac{150}{37.5} = 4$.