If a charge $q$ is placed in a medium of dielectric constant $K$ instead of vacuum, the Coulomb force between $q$ and another charge $q'$ at the same separation is

1. In a medium of permittivity $\varepsilon$, Coulomb's law becomes $F = \dfrac{q q'}{4\pi\varepsilon\,r^2}$. 2. The medium permittivity is $\varepsilon = K\varepsilon_0$, where $K$ (also written $\kappa$ or $\varepsilon_r$) is the DIMENSIONLESS dielectric constant. 3. So $F_\text{medium} = \dfrac{F_\text{vacuum}}{K}$. The presence of a dielectric REDUCES the force by a factor of $K$ (typically $K > 1$). 4. Intuition: the dielectric polarises and partially screens the charges from each other. 5. Examples: water has $K \approx 80$, so Coulomb forces in water are 80× weaker than in vacuum — the reason salts dissociate easily in water. 6. Option A reverses the effect. Option C is wrong — force IS affected. Option D over-reduces. _Source: NCERT Class 12 Physics Part 1, Ch 1, §1.5 (Coulomb's Law in a medium, footnote/discussion), p. 7–8._