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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
A$K$ times larger
B$1/K$ times smaller
CUnchanged — dielectric constant affects only capacitance, not force
D$K^2$ times smaller
Answer & Solution
Correct answer: B. $1/K$ times smaller
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._
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