Does the Coulomb's force depend on the configuration of the medium?

Question

I know that when two charges are placed in a dielectric medium (say water) then the force between the charges get reduced and the force equation is given by :-

$$F=\frac{q_1q_2}{(4π \epsilon_o \epsilon_r) r^2} $$

But I want to ask whether this force also depends on the configuration of the charges inside the medium or not i.e. is the force on the charges different in the two cases given in the figure ?

Note :- The figure is not drawn to scale and assume the distance between the two charges to be same in both the figures. The rectangle represents the region with permittivity $\epsilon_o \epsilon_r$.

Answer 1

Assuming that the rectangle represents the region occupied by the dielectric, the answer is yes; the force depends not only on the distance between the two charges but also on their distance from the boundary of the dielectric region. The physical reason is that at the border between two media with a different dielectric constant, the presence of each charge induces a bound polarization charge modifying the total electric field.

A quantitative treatment of the problem is not trivial. Basically, it can be obtained by using the method of images. The interaction energy between the two charges can be found here (formula ($57$), although that paper derives the formula in a more general (and complex) way. (Warning: I did not have time to check the correctness of the final formula, but the method is sound). The interesting fact is that the presence of the dielectric boundary with the effect of bound charges can even reverse the sign of the forces.