Yes, but it is much more complicated. The molecule does not have to have a dipole, in which case the (static) dielectric constant (relative permittivity) depends on the molecule's polarisability and if it does have a dipole it depends on both factors. The equation is $\displaystyle \epsilon_r=\frac{1+2b}{1-b}$ where $b=\left(\alpha+\mu^2/(k_BT)\right)N/(3\epsilon_0)$ and $N=N_A\rho/M$ where $\alpha$ is polarisability, $\mu$ dipole, $\rho$ density,$M$ molar mass, $\epsilon_0$ permittivity of free space. See Atkins, & Friedman, 'Molecular Quantum Mechanics'. The dielectric constant also depends on frequency of the applied electric field, but the value quoted is normally the static value measured by comparing the capacity of a capacitor with and without the solvent. The name relative permittivity arises from the potential a charge produces. In a vacuum at a distance $r$ this is $\displaystyle \phi=\frac{q}{4\pi\epsilon_0 r}$, in a dielectric medium of permittivity $\epsilon$ this becomes $\displaystyle \phi'=\frac{q}{4\pi\epsilon r}$ and $\epsilon_r=\epsilon/\epsilon_0$.
As an aside you can see why a polar molecule is generally insoluble in a non-polar solvent. If the permittivity is small (say hexane) the electric potential of a charge is higher at any given $r$ than for a solvent with a large $\epsilon$, such as acetonitrile. This means that polar molecules are less soluble in low dielectric solvents than they are in high dielectric ones.
