Question

The force between two charges situated in air is f . The force between the same charges if distance between them is reduced to half and they are situated in a medium having dielwctric constant 4 is

Answer 1

Answer:

F = q₁q₂/4π∈₀r²

permittivity of a medium ∈ = K∈₀ , where K is dielectric constant

In the new case,

r' = r/2

∈ = 4∈₀

F' = q₁q₂/4π*4∈₀(r/2)² = q₁q₂/4π∈₀r²

The new force F' = F

Answer 2

The force remains the same i.e., f.

The force between two charges situated in air is f.

we have to find the force between the same charges if distance between them is reduced to half and they are situated in a medium having dielectric constant 4.

Concept : Coulomb force for free medium,

$F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Coulomb force for a medium of dielectric constant K,

$F'=\frac{1}{4\pi\epsilon_0K}\frac{q_1q_2}{r^2}=\frac{F}{K}$

Now let's solve the problem.

1. distance between them is reduced to half, so the force between them becomes four times of its initial value. i.e., 4f  [ cause force is inversely proportional to square of radius]
2. now they are situated in a medium having dielectric constant K = 4. so the force, f' = 4f/K = 4f/4 = f

Therefore the force acting between them remains constant, if distance between charges is reduced to half and they are situated in a medium having dielectric constant 4.

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