in a parallel plate capacitor the separation between the plates is 3mm with air between them. now a 1mm thick layer of a material of dielectric constant 2 is introduced between the plates due to which capacitance increases. In order to bring the capacitance to the original value, the separation between the plates should be made?
To solve this problem, we need to use the formula for the capacitance of a parallel plate capacitor:
C = (ε₀ * A) / d,
where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.
Given:
Initial separation, d₁ = 3 mm = 0.003 m
Initial permittivity, ε₁ = ε₀ = 8.854 x 10⁻¹² F/m (for air)
Thickness of the dielectric layer, t = 1 mm = 0.001 m
Dielectric constant, κ = 2
We can find the new capacitance, C₂, using the formula for a capacitor with a dielectric material:
C₂ = (κ * ε₀ * A) / (d₁ + t)
To bring the capacitance back to the original value, we need to find the new separation, d₂.
C₁ = (ε₁ * A) / d₂
Since the capacitance needs to be the same, C₁ = C₂. We can set the two equations equal to each other and solve for d₂:
(ε₁ * A) / d₂ = (κ * ε₀ * A) / (d₁ + t)
Simplifying the equation:
d₂ = d₁ * (1 + t/d₁) / κ
Substituting the values:
d₂ = 0.003 * (1 + 0.001/0.003) / 2
d₂ = 0.0015 m = 1.5 mm
Therefore, to bring the capacitance back to the original value, the separation between the plates should be 1.5 mm.