Question

What is the capacitance of a parallel plate capacitor with plates of area 5.00 m^2 separated by 0.100 mm of Teflon?

a) 4.48 × 10^(-11) F
b) 8.85 × 10^(-12) F
c) 2.24 × 10^(-11) F
d) 1.12 × 10^(-11) F

Answer 1

The capacitance of a parallel plate capacitor with plates of area 5.00 m^2 separated by 0.100 mm of Teflon is approximately 1.05 × 10^(-11) F.

Step-by-step explanation:

The capacitance of a parallel plate capacitor can be calculated using the equation:

C = εA/d

Where C is the capacitance, ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates.

In this case, the area of the plates is 5.00 m2 and the distance between them is 0.100 mm. Since Teflon is the material between the plates, we can use the permittivity of Teflon which is approximately 2.1. Plugging these values into the equation, we get:

C = (2.1)(5.00 m2)/(0.100 mm)

Simplifying, we get:

C = 1.05 × 10-11 F

Therefore, the capacitance of the parallel plate capacitor is approximately 1.05 × 10-11 F.