What will be the capacitance of a parallel plate capacitor with air be...
Capacitance of a Parallel Plate Capacitor with Air
The given capacitance of the parallel plate capacitor with air is 8pF.
Formula for capacitance of a parallel plate capacitor with air is given as:
C = ε₀A/d
Where,
- C = Capacitance of parallel plate capacitor
- ε₀ = Permittivity of free space = 8.85 x 10^-12 F/m
- A = Area of one plate of the capacitor
- d = Distance between the plates
Substituting the given values in the formula, we get:
C = ε₀A/d = 8.85 x 10^-12 x A/d
C = 8pF = 8 x 10^-12 F
So, 8 x 10^-12 F = 8.85 x 10^-12 x A/d
A/d = 8 x 10^-12 / 8.85 x 10^-12 = 0.9034
Therefore, Area of one plate of the capacitor divided by the distance between the plates is 0.9034.
Capacitance of the Capacitor if Distance Between the Plates is Reduced to Half
When the distance between the plates is reduced to half, the capacitance of the parallel plate capacitor with air is given as:
C' = ε₀A/d'
Where,
- C' = New capacitance of the parallel plate capacitor
- ε₀ = Permittivity of free space = 8.85 x 10^-12 F/m
- A = Area of one plate of the capacitor
- d' = New distance between the plates = d/2
Substituting the given values in the formula, we get:
C' = ε₀A/d' = 8.85 x 10^-12 x A/(d/2) = 2 x 8.85 x 10^-12 x A/d
C' = 2C = 2 x 8pF = 16pF
Therefore, the new capacitance of the parallel plate capacitor with air when the distance between the plates is reduced to half is 16pF.
Capacitance of the Capacitor if Space Between the Plates is Filled with a Substance of Dielectric Constant 6
When the space between the plates is filled with a substance of dielectric constant 6, the capacitance of the parallel plate capacitor is given as:
C'' = εKA/d
Where,
- C'' = New capacitance of the parallel plate capacitor
- ε = Permittivity of the substance between the plates
- K = Dielectric constant of the substance between the plates = 6
- A = Area of one plate of the capacitor
- d = Distance between