To find the capacitance of a parallel plate capacitor with a Teflon dielectric, we can use the formula:
\[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]
where:
- \( C \) is the capacitance,
- \( \varepsilon_0 \) is the vacuum permittivity (also known as the electric constant),
- \( \varepsilon_r \) is the relative permittivity (or dielectric constant) of Teflon,
- \( A \) is the area of one plate, and
- \( d \) is the separation between the plates.
The vacuum permittivity \( \varepsilon_0 \) has a value of approximately \( 8.854 \times 10^{-12} \, \text{F/m} \) (farads per meter).
Unfortunately, you haven't provided the relative permittivity \( \varepsilon_r \) of Teflon, so I will assume a typical value. The dielectric constant of Teflon is often in the range of 2.0 to 2.1 for most frequencies of interest in capacitor applications. I will assume \( \varepsilon_r = 2.1 \) here, but please use the correct value if it's provided or differs.
The area \( A \) should be in square meters (m²) for the formula to work since the other constants are in SI units. Since we have the area in cm², let's convert it to m²:
\[ A = 7.00 \, \text{cm}^2 \times \left( \frac{1 \, \text{m}}{100 \, \text{cm}} \right)^2 = 7.00 \times 10^{-4} \, \text{m}^2 \]
The separation \( d \) should also be in meters. Again, let's convert mm to m:
\[ d = 0.150 \, \text{mm} \times \left( \frac{1 \, \text{m}}{1000 \, \text{mm}} \right) = 0.150 \times 10^{-3} \, \text{m} \]
Now, we have all the values needed to calculate the capacitance:
\[ C = \frac{(8.854 \times 10^{-12} \, \text{F/m}) \times 2.1 \times (7.00 \times 10^{-4} \, \text{m}^2)}{0.150 \times 10^{-3} \, \text{m}} \]
\[ C = \frac{(1.85934 \times 10^{-12} \, \text{F/m}^2) \times (7.00 \times 10^{-4} \, \text{m}^2)}{0.150 \times 10^{-3} \, \text{m}} \]
\[ C = \frac{1.301538 \times 10^{-12} \, \text{F}}{0.150 \times 10^{-3}} \]
\[ C = 8.67692 \times 10^{-9} \, \text{F} \]
\[ C \approx 8.68 \times 10^{-9} \, \text{F} \]
So, the capacitance of the parallel plate capacitor with Teflon as the dielectric material is approximately \( 8.68 \) nF (nanofarads).
