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. Increase $K$ times
C. Remains unchanged
D. Becomes $\dfrac{1}{{{k^2}}}$ times

Answer 1

We use the Coulombs law of force of attraction to solve this problem. First, we write the formula for the force of attraction in air and then replace the dielectric constant of air with the dielectric constant of the medium to find the maximum force of attraction.

Formula used:
Coulomb’s law of attraction ${F_a} = \dfrac{1}{{4\pi {\varepsilon _0}K}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Force of attraction is represented by ${F_a}$.
Charges of particles one and two are represented by ${q_1},{q_2}$.
Distance between the two particles is represented by $r$.
Dielectric constant is represented by $K$.

Complete step by step answer:
Force of attraction between particles in the air is given as,
${F_a} = \dfrac{1}{{4\pi {\varepsilon _0}K}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
The dielectric constant of air is $1$
The force of attraction air becomes ,
${F_{a - air}} = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here, ${F_{a - air}}$ represents the force of attraction in air.

Force of attraction between particles in the dielectric medium is,
${F_{a - die}} = \dfrac{1}{{4\pi {\varepsilon _0}K}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Comparing the two equations we get,
${F_{a - die}} = \dfrac{1}{{4\pi {\varepsilon _0}K}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}\\\ \therefore{F_{a - die}}= \dfrac{{{F_{a - air}}}}{K}$
Here, ${F_{a - die}}$ represents the force of attraction in presence of a dielectric medium.

Hence the force becomes $\dfrac{{{F_{a - air}}}}{K}$ and decreases by $K$ times.Thus,option A is the correct answer.

Additional information:
Dielectric medium is used in capacitors between its plates to decrease the electric field. This reduces the voltage, thus helps to increase the capacitance. The most widely used media are glass, paper, ceramic, etc.

Note: The force of the dielectric constant is inversely proportional to the force of attraction. Thus, with an increase in $K$ value the force of attraction decreases and vice versa. The same applies to the force of repulsion. Dielectric media are those media that can sustain a static electric field within it. This causes the conductivity of the medium to reduce and makes it slightly insulating. This is the reason why media with dielectric constants have a weaker force of attraction or repulsion.