Answer:
34.7 nF
Explanation:
The capacitance of a parallel-plate capacitor is given by:
[tex]C=\frac{\epsilon_0 \epsilon_r A}{d}[/tex]
where
[tex]\epsilon_0 = 8.85 \cdot 10^{-12} C^2 N^{-1} m^{-2}[/tex] is the permittivity of free space
[tex]\epsilon_r = 9[/tex] is the relative permittivity of the material between the plates
[tex]A = 1.7 m^2[/tex] is the area of the plates
[tex]d=3.9 mm=0.0039 m[/tex] is the distance between the plates
Substituting into the equation, we find:
[tex]C=\frac{(8.85 \cdot 10^{-12})(9)(1.7 m^2)}{(0.0039 m)}=3.47\cdot 10^{-8} F=34.7 nF[/tex]
