A parallel-plate capacitor has a capacitance of 2.1 µF with air between the plates. The capacitor is connected to a 24 V battery and charged. The battery is then removed. When a dielectric is placed between the plates, a potential difference of 2.8 V is measured across the plates. What is the dielectric constant of the material?
In an isolated instance the voltage goes down by a factor of K.
24 v / k = 2.8 v
k = 8.6 (rounded to sig figs)
To find the dielectric constant of the material, we need to use the following formula:
C = ε₀ * εᵣ * A / d
Where:
C is the capacitance of the capacitor,
ε₀ is the permittivity of free space (vacuum) = 8.854 x 10^(-12) F/m,
εᵣ is the dielectric constant of the material between the plates,
A is the area of the plates,
d is the distance between the plates.
We are given that the capacitance C is 2.1 µF, which is equal to 2.1 x 10^(-6) F.
We also know that the initial capacitance with air between the plates is given by:
C = ε₀ * A / d
By rearranging the formula, we can solve for the area A:
A = C * d / ε₀
Substituting the given values, we can calculate the area:
A = (2.1 x 10^(-6) F) * (1 m) / (8.854 x 10^(-12) F/m) = 2.373 x 10^(-4) m^2
Now, let's find the new capacitance C' when the dielectric is inserted and a potential difference of 2.8 V is measured across the plates.
C' = ε₀ * εᵣ * A / d
We can rearrange the formula to solve for the dielectric constant εᵣ:
εᵣ = C' * d / (ε₀ * A)
Substituting the given values, we can calculate the dielectric constant:
εᵣ = (2.1 x 10^(-6) F) * (1 m) / ((8.854 x 10^(-12) F/m) * (2.373 x 10^(-4) m^2)) ≈ 11.48
Therefore, the dielectric constant of the material is approximately 11.48.