Answer
$C = 1.14 \times 10^{-10} F$
Work Step by Step
Given:
Consider a parallel-plate capacitor having an area of 3225 square millimeters (5 $in^2$), a plate separation of 1 mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates.
Required:
capacitance of this capacitor
Solution:
Using a combination of Equations 18.26 and 18.27:
$C = \frac{εA}{l} = \frac{ε_{r}ε_{0}A}{l} = \frac{(4)(8.85 \times 10^{-12} F/m)(3225 mm^{2})(1 m^{2}/10^{6}mm^{2})}{(1 \times 10^{-3} m)} = 1.14 \times 10^{-10} F$