How is Coulomb’s constant affected when it is not in a vacuum permittivity?

According to Gauss' law, the radial electric field due to a point charge $q$ in a linear, uniform medium with relative permittivity $epsilon_r$ is

$$E(r) = frac{1}{4piepsilon_repsilon_0} frac{q}{r^2} = frac{k}{epsilon_r} frac{q}{r^2} $$

where $r$ is the distance from the point charge and $k$ is Coulomb's constant. The force with which two charges $q_1$ and $q_2$ repel each other can similarly be written as

$$ F(r)=frac{1}{4piepsilon_repsilon_0} frac{q_1q_2}{r^2} = frac{k}{epsilon_r} frac{q_1q_2}{r^2}. $$

You see that the "effective" Coulomb's constant in this medium is $k/epsilon_r$. You can intuitively view this as being a result of the polarization of the dielectric partially shielding the electric field due to a free charge.