Answer:
The capacitance of your capacitor is 5.476 x 10⁻⁵ μF
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
[tex]C = \frac{\epsilon A}{d}[/tex]
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
[tex]C = \frac{8.85*10^{-12} *0.02011}{0.00325} = 5.476 * 10^{-11} \ F = \ 5.476 * 10^{-5} \mu F[/tex]
Therefore, the capacitance of your capacitor is 5.476 x 10⁻⁵ μF
