When air is replaced by a dielectric medium of constant K. The maximum force of attraction between two charges separated by a distance:

Question

When air is replaced by a dielectric medium of constant K. The maximum force of attraction between two charges separated by a distance: (A) increases K-1 times (B) increases K times (C) decreases K times (D) remains constant

Tardigrade Admin · Accepted Answer

In air the force of attraction between two charges is

F_{a} = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r ^{2}}

In dielectric medium the force of attraction between two charges is

F_{d} =

\frac{1}{4 π K ϵ _{0}} \frac{q _{1} q _{2}}{r ^{2}} = \frac{F _{a}}{K}

Thus, force decreases K-times.

Q. When air is replaced by a dielectric medium of constant $K$ . The maximum force of attraction between two charges separated by a distance:

increases $K^{- 1}$ times

increases K times

decreases K times

remains constant

In air the force of attraction between two charges is $F_{a} = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r ^{2}}$
In dielectric medium the force of attraction between two charges is $F_{d} =$
$\frac{1}{4 π K ϵ _{0}} \frac{q _{1} q _{2}}{r ^{2}} = \frac{F _{a}}{K}$
Thus, force decreases K-times.

Q. When air is replaced by a dielectric medium of constant K. The maximum force of attraction between two charges separated by a distance:

increases K−1 times