The force between two charges in a medium is given by Coulomb's law modified for the medium:
\[ F' = \frac{1}{4\pi \varepsilon\cdot K} \cdot \frac{q_1 \cdot q_2}{r^2}, \]
where \( F' \) is the force in the medium, \( \varepsilon \) is the permittivity of the medium, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the separation distance.
Since \( F' = 4F \) in this case, you can set up the equation:
\[ 4F = \frac{1}{4\pi \varepsilon\cdot 60} \cdot \frac{q_1 \cdot q_2}{R^2}, \]
where \( R \) is the new separation distance in the medium.
Solving for \( R \):
\[ R = \sqrt{\frac{q_1 \cdot q_2}{60\varepsilon\pi F}}. \]
Please provide the values of the charges (\( q_1 \) and \( q_2 \)) to calculate the distance \( R \).
Let \(F) be the force when charges was in air.
\[ F = \frac{1}{4\pi \varepsilon} \cdot \frac{q_1 \cdot q_2}{R^2}, \]
\[\{frac{F'}{F}} = \frac{60\cdot\(R)}{r}]
]
