When air is replaced by a dielectric medium of force constant K, the maximum force of attraction between two charges, separated by a distance ___ ?

When air is replaced by a dielectric medium of force constant K, the maximum force of attraction between two charges, separated by a distance ___ ?

(a) decreases K-times

(b) increases K-times

(c) remains unchanged

(d) becomes \( {\large \frac{1}{\mathbf{\mathrm{K}}^2}} \)times

Answer : (a) decreases K-times

The electrostatic force between two point charges $q_{\mathrm{1 }}$and $q_{\mathrm{2 }}$, separated by a distance $d$, in any medium is :\[ F = \frac{1}{4\pi\varepsilon} \cdot \frac{q_1 q_2}{d^2} \]

Relative permittivity of the dielectric medium, \[ \varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = K \]

\( \varepsilon_0 \): Permittivity of free space (vacuum)
\( \varepsilon_r \): Relative permittivity of the medium
\( K \): Dielectric constant (same as \( \varepsilon_r \))

Force in vacuum (air):

\[ F_1 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{d^2} \]

Force in a dielectric medium:

\[ F_2 = \frac{1}{4\pi \varepsilon} \cdot \frac{q_1 q_2}{d^2} \]

Taking the ratio of the two forces:

\[ \frac{F_1}{F_2} = \frac{\varepsilon}{\varepsilon_0} = \varepsilon_r = K \]

Therefore:

\[ F_2 = \frac{F_1}{K} \]

So the Force decreases by K times.