Question

on introducing a dielectric material between two positive charges situated in air, the repulsive force between them will
(a) increase
(b) decrease
(c) remain same
​

Answer 1

Answer:

The repulsive force decreases.

Explanation:

According to coulomb's law,  force of repulsion between two positive charge is given by $F=\frac{q_1q_2}{4\pi \epsilon_0 r^2}$, where $\epsilon_0$ is the permittivity of free space.

When a dielectric  material is introduced between two charge, the force of repulsion is given by  $F'=\frac{q_1q_2}{4\pi \epsilon r^2}$, where  $\epsilon$ is the permittivity of medium.

The relation between  dielectric constant k and permittivity is given as $k=\frac{\epsilon}{\epsilon_0}$

⇒$\epsilon=k\epsilon_0$

putting the value in the above formula we get $F'=\frac{q_1q_2}{4\pi k \epsilon_0 r^2}=\frac{F}{k}$.

Hence the Force decreases by a factor of k when a dielectric material is introduced in between two charges. option b is correct