A parallel-plate capacitor has plates measuring 10cm×10cm, spaced 1 mm apart. The dielectric is mica with a relative permittivity of 5.0, and the permittivity of a vacuum is 8.85 ×10-12F/m. Which of these choices is closest to the actual capacitance value?Select one:a.5 pFb.50 pFc.500 pFd.5000 pF

The capacitance (C) of a parallel-plate capacitor is given by the formula:

C = εr * ε0 * (A/d)

where:

• εr is the relative permittivity of the dielectric material (dimensionless),
• ε0 is the permittivity of free space (8.85 × 10^-12 F/m),
• A is the area of one of the plates (in square meters), and
• d is the separation between the plates (in meters).

Given that the plates measure 10 cm × 10 cm (which is 0.1 m × 0.1 m = 0.01 m^2), the separation is 1 mm (which is 0.001 m), the relative permittivity of mica is 5.0, and the permittivity of free space is 8.85 × 10^-12 F/m, we can substitute these values into the formula to find the capacitance:

C = 5.0 * 8.85 × 10^-12 F/m * (0.01 m^2 / 0.001 m) = 4.425 × 10^-10 F.

Since 1 pF = 10^-12 F, we can express this capacitance in picofarads:

C = 4.425 × 10^-10 F * (1 pF / 10^-12 F) = 442.5 pF.

So, the closest answer is:

d. 500 pF.