An air-filled parallel plate capacitor has a plate separation of b = 0.58 cm, and an area A = 280 mm2. Find its capacitance
In response to drwls and bobpursley: most of the time students are posting these questions because they have absolutely no idea where to begin. I can say this from experience. We don't want the answers. We would like to know and understand how to do the problem so we can do it again in the future. No everyone is blessed with a good understanding of physics.
To find the capacitance of an air-filled parallel plate capacitor, you 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 the plates, and d is the separation between the plates.
The permittivity of free space, ε₀, has a constant value of approximately 8.85 x 10⁻¹² F/m.
Now, let's calculate the capacitance step by step.
1. Convert the plate separation from centimeters (cm) to meters (m):
b = 0.58 cm = 0.58 x 10⁻² m
2. Convert the plate area from square millimeters (mm²) to square meters (m²):
A = 280 mm² = 280 x 10⁻⁶ m²
3. Substitute the values into the capacitance formula:
C = (ε₀ * A) / d
= (8.85 x 10⁻¹² F/m) * (280 x 10⁻⁶ m²) / (0.58 x 10⁻² m)
4. Simplify the expression:
C = (8.85 x 10⁻¹² F/m) * (280 x 10⁻⁶ m²) / (0.58 x 10⁻² m)
= (8.85 x 10⁻¹² F) * (280 x 10⁻⁶) / (0.58 x 10⁻²)
= (8.85 x 280 / 0.58) x 10⁻¹² / (10⁻²)
= (2458 / 0.58) x 10⁻¹² / 10⁻²
= 4234.5 x 10⁻¹² / 10⁻²
= 423.45 x 10⁻¹² F
So, the capacitance of the air-filled parallel plate capacitor is approximately 423.45 picofarads (pF).