Answer:
The capacitance of your capacitor is 5.476 x 10⁻⁵ μF
Step-by-step explanation:
Given;
diameter of the aluminum pie plates = 16 cm = 0.16 m
separation distance, d = 3.25 mm = 0.00325 m
voltage across the parallel plates = 6 V
![C = (\epsilon A)/(d)](https://img.qammunity.org/2021/formulas/physics/college/vbpsfryqiugwt7bwp42uxx3dv02nn5ir2n.png)
where;
C is the capacitance of your capacitor
ε is the permittivity of free space = 8.85 x 10⁻¹² (F/m)
d is separation distance
A is the area of the plate = ¹/₄ (πd²) = 0.25 (π x 0.16²) = 0.02011 m²
![C = (8.85*10^(-12) *0.02011)/(0.00325) = 5.476 * 10^(-11) \ F = \ 5.476 * 10^(-5) \mu F](https://img.qammunity.org/2021/formulas/physics/college/vxxcojkn8jy2mn5mj56gy87v9ysau3eovk.png)
Therefore, the capacitance of your capacitor is 5.476 x 10⁻⁵ μF