Answer: 32pF
Explanation:
The capacitance [tex]C[/tex] of a parallel plate capacitor is given by the following equation:
[tex]C=\epsilon_{o}\frac{A}{d}[/tex]
Where:
[tex]\epsilon_{o}=8.85(10)^{-12} \frac{F}{m}[/tex] is the electric constant when the dielectric material is vacuum
[tex]A=6(10)^{-2} m^{2}[/tex] is the area of each plate
[tex]d=4 cm=0.04 m[/tex] is the separation distance between both plates
[tex]C=8.85(10)^{-12} \frac{F}{m}\frac{6(10)^{-2} m^{2}}{0.04 m}[/tex]
[tex]C=1.3275(10)^{-11} F[/tex] This is the capacitance when the dielectric is vacuum.
Now, we can calculate the capacitance [tex]C_{n}[/tex], when we have a dielectric material with a dielectric constant [tex]\epsilon=2.4[/tex]:
[tex]C_{n}=\epsilon C[/tex]
[tex]C_{n}=2.4 (1.3275(10)^{-11} F)[/tex]
Finally:
[tex]C_{n}=3.186 (10)^{-11} F=31.86(10)^{-12} F \approx 32 pF[/tex]
