Question

Force between two point charges q1 and q2 placed in vacuum at ‘r’ cm apart is F. Force between them when placed in a medium having dielectric K= 5 at ‘r/5’ cm apart will be:

Options

• F/25

• 5F

• F/5

• 25F

Answer 1

5F

Explanation:

Given: K = 5

New distance (r') = `r/5`

According to Coulomb’s Law, the electrostatic force (F) between two point charges q₁ and q₂ separated by distance r in a vacuum is:

F = `1/(4 pi epsilon_0) (q_1 q_2)/r^2`    ...(i)

The electrostatic force acting between two point charges in a medium of dielectric constant K is:

F' = `1/(4 pi K epsilon_0) (q_1 q_2)/((r')^2)`

= `1/(4 pi (5) epsilon_0) * (q_1 q_2)/((r/5)^2)`

= `1/(4 pi * 5 epsilon_0) * (q_1 q_2)/(r^2/25)`

= `25/5 * (1/(4 pi epsilon_0) * (q_1 q_2)/r^2)`

= `5 * (1/(4 pi epsilon_0) * (q_1 q_2)/r^2)`    ...[From equation (i)]

= 5F

∴ The force between the point charges when placed in a medium with dielectric constant K = 5 at a distance of r/5 cm will be 5F.