Final answer:
The capacitance of a parallel plate capacitor with plates of area 0.000525 m² and separated by 0.12 mm of Teflon, which has a dielectric constant of 2.1, can be calculated using the formula for capacitance with a dielectric.
Step-by-step explanation:
The student is asking for the capacitance of a parallel plate capacitor with specific dimensions and a Teflon dielectric. The formula for capacitance (C) with a dielectric present is C = ε0εrA/d, where ε0 is the vacuum permittivity (8.85 x 10-12 F/m), εr is the dielectric constant of Teflon (2.1), A is the area of the plates (0.000525 m²), and d is the separation distance (0.12 mm or 0.12 x 10-3 m). Plugging in these values, we find the capacitance of the capacitor.
The capacitance of a parallel plate capacitor can be calculated using the formula C = (ε0 * εr * A) / d, where C is the capacitance, ε0 is the permittivity of free space (8.85 x 10^-12 F/m), εr is the relative permittivity (dielectric constant) of the material, A is the area of the plates, and d is the distance between the plates.
In this case, the area of the plates is given as 5.00 m² and the distance between the plates is 0.100 mm (=0.0001 m). The relative permittivity (dielectric constant) of Teflon is given as 2.1.
Substituting these values into the formula, we get: C = (8.85 x 10^-12 F/m * 2.1 * 5.00 m²) / 0.0001 m = 0.921 F.