When air is replaced by a dielectric medium of dielectric constant \( k \), the force of attraction between the two charges, keeping the distance between them same, will
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → decrease by a factor of \( k \).
By Coulomb’s law in a medium, the permittivity \( \varepsilon_0 \) of free space is replaced by \( \varepsilon = k\varepsilon_0 \), where \( k \) is the dielectric constant of the medium.
\( \displaystyle F_{\mathrm{air}} = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r^{2}} \)
\( \displaystyle F_{\mathrm{medium}} = \frac{1}{4\pi k\varepsilon_0}\,\frac{q_1q_2}{r^{2}} = \frac{F_{\mathrm{air}}}{k} \)
Since \( k \gt 1 \) for every dielectric, the force decreases by the factor k. The molecules of the medium are polarised, and their induced charges partly shield the two charges from each other.