A parallel-plate capacitor is made from two plates 12.0cm on each side and 4.50mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. An 18.0 V battery is connected across the plates.
What is the capacitance of this combination.
So I was using the formula C=(epsilon x area)/d
I'm not getting the right answer
Of course you are not getting the right answer.
This is two parallel plate capacitors in parallel. Figure C for each, then add them.
To calculate the capacitance of the combination, you need to consider the two regions separately: the air-filled region and the Plexiglas-filled region. Then, you can add their capacitance values to obtain the total capacitance.
First, let's calculate the capacitance for the air-filled region:
Given:
Area (A) = (12.0 cm)^2 = 144.0 cm^2 = 0.0144 m^2
Distance between the plates (d) = 4.50 mm = 0.0045 m
Using the formula C = (ε₀ * A) / d, where ε₀ is the permittivity of free space (vacuum), which is approximately 8.85 × 10^-12 F/m, we have:
C₁ = (8.85 × 10^-12 F/m * 0.0144 m^2) / 0.0045 m
C₁ ≈ 2.832 × 10^-11 F
Next, let's calculate the capacitance for the Plexiglas-filled region:
Given:
Dielectric constant (κ) = 3.40
Using the formula C = κ * (ε₀ * A) / d, we have:
C₂ = 3.40 * (8.85 × 10^-12 F/m * 0.0144 m^2) / 0.0045 m
C₂ ≈ 8.589 × 10^-11 F
Finally, add the capacitance values for both regions to obtain the total capacitance:
C_total = C₁ + C₂
C_total = 2.832 × 10^-11 F + 8.589 × 10^-11 F
C_total ≈ 1.3421 × 10^-10 F
Therefore, the capacitance of this combination is approximately 1.3421 × 10^-10 Farads.
To find the capacitance of the combination of a parallel-plate capacitor with half air and half Plexiglas as the dielectric, you need to consider the different dielectric constants and the total effective distance between the plates.
The formula you mentioned, C = (ϵ x area) / d, is correct. However, since the dielectric material is different in each half, you need to consider the effective dielectric constant and effective distance. Let's break down the calculation step-by-step:
1. Calculate the capacitance for the air-filled half:
Since the air has a dielectric constant of 1, the effective dielectric constant for the air-filled half is also 1. The distance between the plates for the air-filled half is half the total distance, which is 4.50mm / 2 = 2.25mm = 0.0225m.
Using the formula C = (ϵ x area) / d, where the area is the area of one plate (12.0cm x 12.0cm = 0.12m x 0.12m = 0.0144 m²), and the dielectric constant ϵ is 1, we get:
C_air = (1 x 0.0144) / 0.0225 ≈ 0.009216 F.
2. Calculate the capacitance for the Plexiglas-filled half:
The effective dielectric constant for the Plexiglas-filled half is given as 3.40. The distance between the plates for the Plexiglas-filled half is also 0.0225 m.
Using the same formula, we get:
C_Plexiglas = (ϵ x area) / d = (3.40 x 0.0144) / 0.0225 ≈ 0.02183 F.
3. Calculate the total capacitance:
Since the two halves are connected in series, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances:
1 / C_total = 1 / C_air + 1 / C_Plexiglas.
Substituting the values, we have:
1 / C_total = 1 / 0.009216 + 1 / 0.02183 ≈ 108.51 + 45.841 ≈ 154.35.
Taking the reciprocal of both sides, we get:
C_total ≈ 1 / 154.35 ≈ 0.00647 F.
Therefore, the capacitance of the combination of the air-filled and Plexiglas-filled parallel-plate capacitor is approximately 0.00647 F.