Answer:
1.6 pF
Explanation:
The capacitance of a parallel-plate capacitor in air is given by:
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of each plate
d is the separation between the plates
In this problem we have:
- Separation between the plates: d = 5.00 mm = 0.005 m
- Area of the plates: [tex]A = 3.00 cm \cdot 3.00 cm = 9.00 cm^2 = 9\cdot 10^{-4}m^2[/tex]
Therefore, the capacitance is
[tex]C=\frac{(8.85\cdot 10^{-12}F/m)(9\cdot 10^{-4} m^2)}{0.005 m}=1.6\cdot 10^{-12} F=1.6 pF[/tex]