Coulomb force between two point charges in a dielectric medium

Question

In a dielectric medium with relative dielectric constant $\epsilon_r$, what is the Coulomb force between two free point charges $q_1$ and $q_2$ at distance $r$? Is it equal to the Coulomb force in vacuum divided by $\epsilon_r$ or $\epsilon_r^2$, i.e., whether the formula is

$$F=\frac{q_1q_2}{4\pi\epsilon_0\epsilon_rr^2}\quad\mbox{or}\quad\frac{q_1q_2}{4\pi\epsilon_0\epsilon_r^2r^2}?$$

I know in a dielectric medium, we have $\nabla^2\phi=-\rho_0/(\epsilon_0\epsilon_r)$. Assuming the medium is infinitely big with no boundary to consider, the Coulomb field generated by either $q_1$ or $q_2$ is reduced by a factor of $\epsilon_r$. But I also know this effect is due to the bound charges $-q_1(1-1/\epsilon_r)$ and $-q_2(1-1/\epsilon_r)$ that surround the free charges $q_1$ and $q_2$, leaving net charges $q_1/\epsilon_r$ and $q_2/\epsilon_r$. So let's say if the $q_1$ and $q_2$ are like charges and I connect them with an insulating rope. Which force formula is correct to use, if I want to calculate the tension in the rope at equilibrium, assuming the medium is a frictionless fluid?

Answer 1

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.

Answer 2

The force analysis of the problem is done thanks to @freecharly. I now work it out using two other methods: virtual work and field energy. Suppose charge $q_1$ is fixed and charge $q_2$ moves along the rope by a small virtual displacement $\delta r$ away from $q_1$. The Coulomb field due to the net charge $q_1/\epsilon_r$ repels $q_2$ and attracts the bound charge $-q_2(1-1/\epsilon_r)$ surrounding $q_2$. When $q_2$ moves by the distance $\delta r$, the bound charge doesn't really move with $q_2$. By definition, bound charge cannot move. What actually happens is that the bound charge at the original position of $q_2$ depolarizes to neutrality, while some new bound charge of the same amount reappears at the new position of $q_2$. So no work is done to the bound charge $-q_2(1-1/\epsilon_r)$ because no bound charge actually moved the distance $\delta r$. Therefore,

$$F\delta r=\frac{q_1q_2}{4\pi\epsilon_0\epsilon_rr^2}\delta r,$$

which means the first formula is correct. Notice that the visionary "displacement" of the bound charges $-q_2(1-1/\epsilon_r)$ is not a real displacement. Therefore no work is done to them.

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The field energy method does not distinguish free charges and bound charges or track how charges move. It uses the energy of capacitors $\,W=\frac{1}{2}CU^2\,$ with $\,C=\epsilon_0\epsilon_rS/d\,$ and $\,E=U/d\,$ to obtain

$$W=\frac{1}{2}\epsilon_0\epsilon_rE^2Sd.$$

Therefore the energy density of an $E$-field in a dielectric medium is greater than the same $E$-field in vacuum by a factor of $\epsilon_r$ due to the polarization of the medium. When the distance between the net charges $q_1/\epsilon_r$ and $q_2/\epsilon_r$ increases by $\delta r$, if these were charges in vacuum, the $E$-field energy would reduce by an amount equal to the work done by the force in the second formula times $\delta r$. But now the $E$-field of the net charges $q_1/\epsilon_r$ and $q_2/\epsilon_r$ are in the medium. Therefore, the work done is $\epsilon_r$ times greater. In formula we have

$$F\delta r=\frac{(q_1/\epsilon_r)(q_2/\epsilon_r)}{4\pi\epsilon_0r^2}\delta r\times\epsilon_r=\frac{q_1q_2}{4\pi\epsilon_0\epsilon_rr^2}\delta r.$$

$E$-fields in dielectric media contain more energy than the same $E$-fields in vacuum by $\epsilon_r$ times.