Question

A parallel plate capacitor has two plates, both have an area of 1.7 m2 and the distance between the plates is 3.9 mm. The relative permittivity of the material between the plates is 9. What is the capacitance value, in units of nanoFarads, for this parallel plate capacitor?
Remember the permittivity of free space is equal to 8.85 x 10-12 C2N-1m-2 and the formula for a parallel plate capacitor is:

Answer 1

Answer:

34.7 nF

Explanation:

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where

$\epsilon_0 = 8.85 \cdot 10^{-12} C^2 N^{-1} m^{-2}$ is the permittivity of free space

$\epsilon_r = 9$ is the relative permittivity of the material between the plates

$A = 1.7 m^2$ is the area of the plates

$d=3.9 mm=0.0039 m$ is the distance between the plates

Substituting into the equation, we find:

$C=\frac{(8.85 \cdot 10^{-12})(9)(1.7 m^2)}{(0.0039 m)}=3.47\cdot 10^{-8} F=34.7 nF$