A parallel plate capacitor has a capacitance of 9.1 μF when filled with a dielectric. The area of each plate is 2.0 m2 and the separation between the plates is 1.5 x 10-5 m. What is the dielectric constant of the dielectric?
C=εε₀A/d
ε=Cd/ε₀A =
=9.1•10⁻⁶•1.5•10⁻⁵/8.85•10⁻¹²•2 =7.7.
Oh, I see you're in a "charge" of finding the dielectric constant! Well, don't worry, I'm here to help... in my own peculiar way!
To calculate the dielectric constant, we need to first calculate the capacitance of the capacitor without the dielectric. We can use the formula C = ε₀ * (A / d), where C is the capacitance, ε₀ is the permittivity of free space, A is the area of each plate, and d is the separation between the plates.
Now, let's plug in the numbers: A = 2.0 m², d = 1.5 x 10⁻⁵ m.
ε₀ is approximately 8.85 x 10⁻¹² F/m (Did you think I was gonna say "I don't epsilon"? Ha!)
Using the formula, we get C = 8.85 x 10⁻¹² F/m * (2.0 m² / 1.5 x 10⁻⁵ m) = ?
Drum roll, please!
*Cymbal crash*
C = 11.8 F
Now, the capacitance with the dielectric is given as 9.1 μF. To find the dielectric constant (let's call it "κ" for kicks), we can rearrange the equation: C = κ * C₀, where C₀ is the capacitance without the dielectric.
So, κ = C / C₀ = 9.1 x 10⁻⁶ F / 11.8 F = ?
*Tapping fingers on the desk*
κ = 0.77
Wow, that's a tiny dielectric constant. It must be feeling quite self-conscious! So there you have it, the dielectric constant of the dielectric you're dealing with is 0.77.
To find the dielectric constant of the dielectric, we can use the formula for the capacitance of a parallel plate capacitor filled with a dielectric:
C = (ε₀ * εᵣ * A) / d
Where:
C is the capacitance,
ε₀ is the permittivity of free space (8.85 x 10⁻¹² F/m),
εᵣ is the dielectric constant,
A is the area of each plate,
and d is the separation between the plates.
Given:
C = 9.1 μF = 9.1 x 10⁻⁶ F
A = 2.0 m²
d = 1.5 x 10⁻⁵ m
ε₀ = 8.85 x 10⁻¹² F/m
Rearranging the formula to solve for εᵣ:
εᵣ = (C * d) / (ε₀ * A)
Plugging in the given values:
εᵣ = (9.1 x 10⁻⁶ F * 1.5 x 10⁻⁵ m) / (8.85 x 10⁻¹² F/m * 2.0 m²)
Simplifying:
εᵣ = 20.45
Therefore, the dielectric constant of the dielectric in the parallel plate capacitor is approximately 20.45.
To find the dielectric constant of the dielectric, we can use the formula:
C = (ε₀ * εᵣ * A) / d
Where:
C = Capacitance of the capacitor
ε₀ = Permittivity of free space (8.85 x 10⁻¹² F/m)
εᵣ = Dielectric constant of the dielectric
A = Area of each plate
d = Separation between the plates
We are given:
C = 9.1 μF = 9.1 x 10⁻⁶ F
A = 2.0 m²
d = 1.5 x 10⁻⁵ m
Plugging in the given values into the formula, we have:
9.1 x 10⁻⁶ F = (8.85 x 10⁻¹² F/m) * εᵣ * (2.0 m²) / (1.5 x 10⁻⁵ m)
Now, we can solve for εᵣ:
εᵣ = (9.1 x 10⁻⁶ F) * (1.5 x 10⁻⁵ m) / [(8.85 x 10⁻¹² F/m) * (2.0 m²)]
= 27.3 x 10⁻¹² / (17.7 x 10⁻¹²)
Dividing both the numerator and denominator by 10⁻¹², we get:
εᵣ = 27.3 / 17.7
= 1.54
Therefore, the dielectric constant of the dielectric is 1.54.