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zhannawk [14.2K]
3 years ago
9

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 pl

ates are separated by a dielectric made of Teflon , which has a dielectric constant of 2.1. What is the capacitance of the capacitor?
Physics
1 answer:
Aleks04 [339]3 years ago
5 0

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

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

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