An air-filled parallel plate capacitor has a plate separation of 0.6 cm, and an area 179 mm2.
In femto-farads (fF), what is the capacitance of this capacitor?
To calculate the capacitance of an air-filled parallel plate capacitor, you can use the equation:
C = (𝜀₀ * 𝐴) / 𝑑
where:
C is the capacitance,
𝜀₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m),
𝐴 is the area of the plates, and
𝑑 is the separation between the plates.
Let's first convert the values given into the appropriate units:
- The plate separation is given as 0.6 cm, which is equal to 0.006 m.
- The area is given as 179 mm², which is equal to 179 x 10^-6 m² (since 1 mm = 10^-3 m).
Now we can substitute the values into the equation:
C = (8.85 x 10^-12 F/m * 179 x 10^-6 m²) / 0.006 m
Simplifying this expression gives us:
C = 24.89 x 10^-18 F
To convert this into femto-farads (fF), we need to multiply the value by 10^15:
C = 24.89 x 10^-18 F * 10^15
C = 24.89 fF (rounded to two decimal places)
Therefore, the capacitance of the air-filled parallel plate capacitor is approximately 24.89 fF.