Two point electric charges interact in air at a distance of 0.4 m with the same force as in a non-conductive

Two point electric charges interact in air at a distance of 0.4 m with the same force as in a non-conductive liquid at a distance of 0.2 m. Determine the dielectric constant of a non-conductive liquid.

According to Coulomb’s law, the force of interaction of point charges

F = k * q1 * q2 / (e * r²),
where
k is a constant coefficient;
q1 and q2 are the magnitudes of charges;
e – dielectric constant:
r is the distance between charges.

Let us denote the dielectric constant of air and liquid through e1, e2.
Let’s make the equation

F1 = F2;
k * q1 * q2 / (e1 * r1²) = k * q1 * q2 / (e2 * r2²).

Reduce and get

e2 = e1 * r1² / r2²
e₂ = 1 * 0.4² / 0.2 ² = 0.16 / 0.04 = 4.

Answer: 4.