Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric of dielectric constant 4 is inserted between these two charges. If the thickness of the slab is r/2, then the force between the charges will becomea)Fb)3/5Fc)4F/9d)F/4e)F/2Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Solution:

Concept: The force between two charges is given by Coulomb's Law as F = (1/4πε₀) (q₁q₂/r²), where ε₀ is the permittivity of vacuum.

When a dielectric is inserted between two charges, the force between them changes. The force becomes F' = (1/4πε₀ε_r) (q₁q₂/r²), where ε_r is the relative permittivity or dielectric constant of the material.

Given: F = force between two identical charges in vacuum, dielectric constant of the slab = 4, thickness of the slab = r/2

To find: The force between the charges when the slab is inserted.

Solution:

We know that F = (1/4πε₀) (q₁q₂/r²)

Let q₁ = q₂ = q

So, F = (1/4πε₀) (q²/r²)

When the slab is inserted, the force between the charges becomes F' = (1/4πε₀ε_r) (q²/r²)

We know that the thickness of the slab is r/2, so the distance between the charges is now 2r.

Therefore, the new force between the charges is F' = (1/4πε₀ε_r) (q²/(2r)²) = (1/16πε₀ε_r) (q²/r²)

So, F'/F = [(1/16πε₀ε_r) (q²/r²)] / [(1/4πε₀) (q²/r²)] = ε_r/4

Given ε_r = 4, so F'/F = 1/4

Therefore, F' = F/4 = (4F/16) = 4F/16 = 4F/9

Hence, the correct option is (c) 4F/9.

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