Answer :
To find the capacitance of an air-filled parallel-plate capacitor, we need to use the formula for the capacitance [tex]C[/tex] of a parallel-plate capacitor:
[tex]C = \frac{\varepsilon_0 \cdot A}{d}[/tex]
where:
- [tex]\varepsilon_0[/tex] is the permittivity of free space, which is approximately [tex]8.854 \times 10^{-12} \text{ F/m}[/tex]
- [tex]A[/tex] is the area of one plate in square meters
- [tex]d[/tex] is the separation between the plates in meters
Given in the problem:
The area [tex]A = 10.08 \text{ cm}^2[/tex]. First, we need to convert this to square meters.
[tex]1 \text{ cm}^2 = 10^{-4} \text{ m}^2[/tex]
[tex]A = 10.08 \times 10^{-4} \text{ m}^2[/tex]The separation [tex]d = 1.57 \text{ mm}[/tex]. We need to convert this to meters.
[tex]1 \text{ mm} = 10^{-3} \text{ m}[/tex]
[tex]d = 1.57 \times 10^{-3} \text{ m}[/tex]
Now plugging these values into the formula:
[tex]C = \frac{8.854 \times 10^{-12} \text{ F/m} \times 10.08 \times 10^{-4} \text{ m}^2}{1.57 \times 10^{-3} \text{ m}}[/tex]
[tex]C \approx \frac{8.91 \times 10^{-15} \text{ F/m}^2}{1.57 \times 10^{-3} \text{ m}}[/tex]
[tex]C \approx 5.68 \times 10^{-12} \text{ F}[/tex]
Therefore, the capacitance is approximately [tex]5.68 \text{ pF}[/tex] (picoFarads). This value represents the ability of the capacitor to store charge per unit voltage across its plates.
