When a medium of dielectric constant K is put between two point charges the force becomes

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Two point charges are placed at separation d in vacuum and the force between them is F. Now a dielectric slab of thickness t=d3 and dielectric constant K is placed between the charges and the force becomes 9F25. Find the value of K.

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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.

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.

How does the force between two point charges change, if the dielectric constant of the medium in which they are kept decreases?

Dielectric constant of a medium is defined as the ratio of forces between two charges placed in vacuum to medium without any change in magnitude of charges and their distance. 

∴             K = F0F

⇒             F = F0K

i.e.,            F∝1k

Therefore, as K decreases there will be an increase in force.