Question

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 $\frac{1}{K^2}$

Answer 1

The correct option is A Decreases $K$-times

The force between two charges is inversely proportional to the dielectric constant of the medium in which charges are kept.

In air $F_{air}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
In medium, $F_m=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$\therefore \frac{F_m}{F_{air}}=\frac{1}{K}\Rightarrow F_m=\frac{F_{air}}{K}$ (decreases K- times)