Question

The rectangular plates in a parallel-plate capacitor are 0.063 m x 5.4 m. A distance of 3.5 x 10^-5m separate the plates. The plates are separated by a dielectric made of Teflon , which has a dielectric constant of 2.1. What is the capacitance of the capacitor?

Answer 1

The capacitance of the capacitor is $1.8\cdot 10^{-9}F$

Explanation:

The capacitance of a parallel-plate capacitor is given by the equation

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

where

k is the dielectric constant of the medium

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the capacitor in this problem, we have:

k = 2.1 is the dielectric constant

$d=3.5\cdot 10^{-5}m$ is the separation between the plates

$A=0.063m \cdot 0.054m = 0.0034 m^2$ (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)

Therefore, the capacitance of the  capacitor is

$C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F$

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