The Coulomb force in a medium with relative dielectric constant $\epsilon_r$ is given by your first equation. Only from this follows the electric field strength of a spherical symmetric free charge $Q$ in the dielectric with $$E=\frac{Q}{4\pi\epsilon_0\epsilon_r r^2} \tag{1}$$ which, with the electric displacement $D=\epsilon_r \epsilon_0 E$, results in the correct Gauss Law $$ \int_{sphere} \epsilon_r \epsilon_0 E da=Q \tag{2}$$ This is equivalent to the differential form of Gauss's Law, the Maxwell equation in a dielectric $$ div (\epsilon_r \epsilon_0 \vec E)=\rho$$ where $\rho$ is the free charge density.
Note added after a comment by Zhouran He: In Coulomb's Law for the electric force $F$ exerted by a free charge $q_1$ on a second (test) charge $q_2$ in a dielectric with relative permittivity $\epsilon_r$, only the charge $q_1$ as the source of the force field can be considered to be reduced by the polarization charges of the dielectric to the $q_1/\epsilon_r$ so that the vacuum Coulomb law can be used with this net charge. Even though the charge $q_2$ is also surrounded by polarization charges, the force $F$ exerted by the net charge $q_1/\epsilon_r$ works on the free charge $q_2$. One can alternatively consider $q_2/\epsilon_r$ to be the net charge exerting the force $F$ on the free (test) charge $q_1$. The free charge $q_2$ sees a net charge $q_1/\epsilon_r$ exerting a force $F$ on it according to Coulombs vacuum law. The polarization charges induced by itself around it don't exert a force on itself. The same reasoning applies with interchanged roles of the charges. Thus the second form of Coulombs Law for a dielectric is correct.
In this video by Flammable Maths, the solution to a similar problem is given.
The only difference is that we just need to include the electrostatic force, besides that the process is exactly the same.
Let's say we have two objects $1$ and $2$ with mass $m_1,m_2$ and charge $q_1,q_2$ respectivey separated by distance $R$ then-
$$\textstyle\displaystyle{F=F_C+F_G=\frac{Gm_1m_2+kq_1q_2}{R^2}}$$
Where $G$ is the Newtonian constant of gravitation and $$\textstyle\displaystyle{k=\frac{1}{4\pi\epsilon_0}}$$
By newton's third law we have $F_{12}=-F_{21}$ so
$$\textstyle\displaystyle{F_{12}=\frac{Gm_1m_2+kq_1q_2}{(r_2-r_1)^2}=m_1\frac{d^2r_1}{dt^2}}$$
$$\textstyle\displaystyle{F_{21}=-\frac{Gm_1m_2+kq_1q_2}{(r_2-r_1)^2}=m_2\frac{d^2r_2}{dt^2}}$$
Where $R=r_2-r_1$
$$\therefore\textstyle\displaystyle{\frac{d^2r_2}{dt^2}-\frac{d^2r_1}{dt^2}}$$
$$\textstyle\displaystyle{=-\frac{Gm_1m_2+kq_1q_2}{(r_2-r_1)^2}\bigg(\frac{1}{m_1}+\frac{1}{m_2}\bigg)}$$
$$\implies\textstyle\displaystyle{\frac{d^2R}{dt^2}=-\frac{\kappa}{R^2}}$$
Now we just need to solve this differential equation-
$$\textstyle\displaystyle{\frac{dv}{dt}=-\frac{\kappa}{R^2}=\frac{dv}{dR}\frac{dR}{dt}}$$
$$\implies\textstyle\displaystyle{-\frac{\kappa}{R^2}=v\frac{dv}{dR}}$$
$$\implies\textstyle\displaystyle{-\kappa\int\frac{1}{R^2}dR=\int vdv}$$
At $t=0$, $R(0)=R_i$ [The initial radius]
$v(0)=0$ [velocity at the beginning]
$$\therefore\textstyle\displaystyle{\int_{0}^{v(t)}vdv=-\kappa\int_{R_i}^{R(t)}R^{-2}dR}$$
$$\implies\textstyle\displaystyle{\frac{v^2}{2}=\kappa\bigg(\frac{1}{R}-\frac{1}{R_i}\bigg)}$$
$$\implies\textstyle\displaystyle{v=\frac{dR}{dt}=\pm\sqrt{2\kappa\bigg(\frac{R_i-R}{R_iR}\bigg)}}$$
$$\implies\textstyle\displaystyle{\int_{0}^{T_c}dt=\pm\int_{R_i}^{0}\frac{1}{\sqrt{2\kappa\bigg(\frac{R_i-R}{R_iR}\bigg)}}dR}$$
$$\implies\textstyle\displaystyle{T_c=\pm\sqrt{\frac{R_i}{2\kappa}}\int_{R_i}^{0}\sqrt{\frac{R}{R_i-R}}dR}$$
Solving the integral is simple, if you would like to see the steps then see here. Noting that time can't be negative, we have-
$$\textstyle\displaystyle{T_c=\frac{\pi}{2}\sqrt{\frac{R^3}{2\kappa}}}$$
Now simply substituting the value for $\kappa$ and $k$ gives us less cleaner formula-
$$\textstyle\displaystyle{T_c=\sqrt{\frac{\pi^3\epsilon_0m_1m_2R^3}{2(m_1+m_2)(4\pi\epsilon_0Gm_1m_2+q_1q_2)}}}$$
Best Answer
From https://www.britannica.com/science/dielectric
Of course saying ‘lower field’ is identical to saying ‘reduced force on a charged particle’. The field from one particle is reduced and the force pushing or pulling on the other particle in response is reduced, by the same proportion (we wouldn’t reduce the fields for each particle, that would be double counting; we just reduce one particle’s field, for either particle as force is same either way, and say, ‘What’s the force on the other charge now?’).
Field is just a way of describing coulomb attraction/repulsion, as different configurations can give same net forces at various points (think of electric field as force per unit charge, force on a unit charge at that location being equivalent to field at that location). It’s useful and tractable.
And the field is reduced because “the quantity of energy stored in an electric field per unit volume of a dielectric medium is greater”.. than the energy stored in the same field in a vacuum. So the fields wouldn’t be the same - that would increase energy not conserve it, just by moving a dielectric into a field.
Field and energy are related by the concept of electric potential, the energy to move a unit-charge through the field, force times distance (actually integral of Fdt), is electric potential, similar to energy in the field. We could say part of the the field is being ‘used’ to keep the dielectric medium polarized, to hold the dipoles that comprise it in alignment with the field, ie to keep those dipoles polarized.
Another way to say all this is that the dielectric constant of air is 1.