# A parallel plate capacitor has a capacitance of 7.0mF when filled with a dielectric. The area of each plate is 1.5m2 and the separation between the plates is 1.0x10-5m. What is the dielectric constant of the dielectric?

## To find the dielectric constant of the dielectric, we can use the formula for the capacitance of a parallel plate capacitor filled with dielectric:

C = ε₀ * εᵣ * A / d

Where:
C = Capacitance
ε₀ = Permittivity of free space (8.85 x 10⁻¹² F/m)
εᵣ = Dielectric constant
A = Area of each plate
d = Separation between the plates

Given:
C = 7.0 mF
A = 1.5 m²
d = 1.0 x 10⁻⁵ m

Now we can rearrange the formula to solve for the dielectric constant:

εᵣ = C * d / (ε₀ * A)

Substituting the given values into the formula:

εᵣ = (7.0 x 10⁻³ F) * (1.0 x 10⁻⁵ m) / (8.85 x 10⁻¹² F/m * 1.5 m²)

Now, let's calculate the dielectric constant:

εᵣ = 4.68 x 10⁵

Therefore, the dielectric constant of the dielectric is approximately 4.68 x 10⁵.

## To find the dielectric constant, we can use the formula for the capacitance of a parallel plate capacitor with a dielectric:

C = ε₀ * εᵣ * A / d

Where:
C is the capacitance of the capacitor
ε₀ is the permittivity of free space (a constant value)
εᵣ is the dielectric constant of the material
A is the area of each plate
d is the separation between the plates

In this case, we are given:
A = 1.5 m² (square meters)
d = 1.0 x 10⁻⁵ m (meters)

To solve for εᵣ, we rearrange the formula:

εᵣ = C * d / (ε₀ * A)

Now, let's plug in the given values and calculate the dielectric constant:

ε₀ = 8.85 x 10⁻¹² F/m (permeability of free space)

εᵣ = (7.0 x 10⁻³ C * 1.0 x 10⁻⁵ m) / (8.85 x 10⁻¹² F/m * 1.5 m²)

Simplifying this expression:

εᵣ = (7.0 x 10⁻³) / (8.85 x 10⁻¹² * 1.5)

εᵣ ≈ 5,028

Therefore, the dielectric constant of the dielectric material is approximately 5,028.