Two charges are placed in air attract each other with force of 1 N. Without changing the distance if they are placed in a medium with dielectric constant 2, what will be the force between them?
A) 1 N
B) 2 N
C) 0.5 N
D) 0.05 N
Answers
Answer:
Option C) 0.5 N. The force between the charges in presence of the dielectric is 0.5 N.
Explanation:
The force of attraction between two charges is given by the coulomb's law.
F = (1/4πε₀) (q₁q₂ / r²)
Where 'F' is the force of interaction between the charges, q₁ and q₂ are the charges separated by a distance 'r', and ε₀ is the permittivity in free space.
If a dielectric is introduced between the charges, the equation will change as:
F' = (1/4πε) (q₁q₂ / r²)
We know relative permittivity, εₙ = ε / ε₀
⇒ ε = εₙε₀
Where ε is the permittivity of the dielectric medium.
∴ F' = (1/4πεₙε₀) (q₁q₂ / r²)
F' = 1/εₙ × (1/4πε₀) (q₁q₂ / r²)
F' = 1/εₙ × F
The value of dielectric constant, εₙ = 2
The force between the charges in free space or air is F = 1 N
∴ The force between the charges in presence of a dielectric medium,
F' = 1/2 × 1
F' = 1/2 = 0.5 N