Question

Find the capacitance (in µF) of a parallel plate capacitor having plates of area 7.00 m2...

Find the capacitance (in µF) of a parallel plate capacitor having plates of area 7.00 m2 that are separated by 0.300 mm of Teflon.

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Answer #1
Concepts and reason

The concept of the capacitance for dielectric constant are used to solve this problem.

Use the parallel-plate capacitance for dielectric constant to calculate the capacitance of Teflon.

Fundamentals

Write the expression of the parallel-plate capacitance for dielectric constant.

C=ke, a

Here, is the permittivity of free space, is the dielectric constant, is the area of the parallel plates and is the distance between two plates.

Draw the diagram of the parallel-plate capacitor with dielectric material between the plates.

Teflon
+
[ [
+
(
+
+
( ]
+
+
( ]
+
+
[
+
[ᎾᎾᎾᎾᎾ C]-
+
( ]
+
+
[ [
+)
+
+
+
( ]
+
( ]
+
+
+
( ]
+
+
[

Understand that, the above figure shows the parallel-plate capacitor with dielectric Teflon among the plates.

Write the expression of the parallel-plate capacitance.

C=ke, a

Substitute for , 8.85x10-12 F/m
for , 7.00 m²
for and 0.300 mm
for .

C =(2.1)(8.85*10-2 F/m) 7.00 m²
= (2.1)(8.85x10-12 F/m) -
7.00 m²
10
0.300 mm 1 mm
m
= 4.34x10-7
l 10° uf
IF
=0.434 uF

Ans:

The capacitance of a parallel plate is 0.434 uF
.

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