The voltage across an air-filled parallel-plate capacitor is measured to be 219.0 V. When a dielectric is inserted and completely fills the space between the plates as in the figure below, the voltage drops to 64.4 V. What is the dielectric constant of the inserted material? Can you identify the dielectric? (b) If the dielectric doesn't completely fill the space between the plates, what could you conclude about the voltage across the plates?
The charge on the capacitor remains the same but the capacitance increases in proportion to the dielectric constant. The product C*V (= Q) remains constant.
In this case, the drop in V by a factor of 3.40 means that C increases by the same factor and that the dielectric constant is 3.40.
That is a typical value for many plastics such as epoxy and polyimide.
To find the dielectric constant of the inserted material, we can 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 (8.85 x 10^-12 F/m)
A is the area of the plates
d is the distance between the plates
In this case, the voltage changed from 219.0 V to 64.4 V, so let's calculate the capacitance before and after the dielectric is inserted.
Before the dielectric is inserted, the voltage is 219.0 V. Assuming the area and distance between the plates remain the same, we can find the capacitance:
C₀ = ε₀ * A / d₀
Where:
C₀ is the initial capacitance
ε₀ is the permittivity of free space (8.85 x 10^-12 F/m)
A is the area of the plates
d₀ is the distance between the plates
Now, after the dielectric is inserted, the voltage drops to 64.4 V. Let's find the capacitance with the dielectric:
C₁ = ε₀ * k * A / d₀
Where:
C₁ is the capacitance with the dielectric
ε₀ is the permittivity of free space (8.85 x 10^-12 F/m)
k is the dielectric constant of the material
A is the area of the plates
d₀ is the distance between the plates
To find the dielectric constant, we can rearrange the equation as:
k = C₁ / C₀
Now, let's calculate the dielectric constant:
k = (ε₀ * k * A / d₀) / (ε₀ * A / d₀)
k = (k * A) / A
k = C₁ / C₀
Given the voltage drops as stated, we have:
C₀ = (ε₀ * A / d₀) = (ε₀ * A / d₁)
Therefore, the dielectric constant, k, is equal to the voltage drop ratio:
k = 64.4 V / 219.0 V
To identify the dielectric, we need to compare the calculated dielectric constant to known material dielectric constants. Different materials have different dielectric constants, so we can refer to a table of dielectric constants to identify the material.
If the dielectric doesn't completely fill the space between the plates, we can conclude that the voltage across the plates would be lower compared to a fully inserted dielectric. The voltage drop is directly proportional to the dielectric's effect on the capacitance. Therefore, if the dielectric doesn't completely fill the space, the capacitance and hence the voltage drop will be lower than if it was fully inserted.
To find the dielectric constant of the inserted material, we can use the formula:
C = ε₀ * εᵣ * A / d
Where:
C is the capacitance of the capacitor,
ε₀ is the permittivity of free space (8.854 x 10⁻¹² F/m),
εᵣ is the dielectric constant of the material,
A is the area of the capacitor plates, and
d is the distance or separation between the plates.
In this case, we have the voltages across the capacitor plates, and we know that the capacitance remains the same when a dielectric is inserted.
First, we can find the capacitance before the dielectric is inserted:
C₁ = ε₀ * A / d₁
Next, we can find the capacitance after the dielectric is inserted:
C₂ = ε₀ * εᵣ * A / d₂
Since the capacitance remains the same, we can equate these two equations:
C₁ = C₂
ε₀ * A / d₁ = ε₀ * εᵣ * A / d₂
The area of the capacitor plates, A, and the permittivity of free space, ε₀, are constants and cancels out on both sides. Rearranging the equation, we get:
εᵣ = d₁ / d₂
Substituting the given values, we have:
εᵣ = 219.0 V / 64.4 V
Simplifying the fraction gives us the dielectric constant:
εᵣ ≈ 3.4
To identify the dielectric material, we need more information such as a known list of dielectric constants for different materials. Comparing the calculated dielectric constant (approximately 3.4) to known values can help determine the material. For example, materials like Teflon, polystyrene, and some ceramics have dielectric constants around this range.
Regarding the second part of your question, if the dielectric material does not completely fill the space between the plates, the effective capacitance of the system will decrease. Consequently, the voltage across the plates will increase. This is because the presence of the dielectric enhances the overall capacitance, thus reducing the voltage drop.