Question

the plates of a parallel-plate capacitor are 3.20 mm apart, and each has an area of 9.92 cm2 . each plate carries a charge of magnitude 4.35×10−8 c . the plates are in vacuum.

Answer 1

The capacitance of the parallel-plate capacitor is approximately 27.45 picofarads (pF).

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

C = ε₀A/d

Where C is the capacitance, ε₀ is the permittivity of free space (a constant value of approximately 8.85 × 10-12 F/m), A is the area of the plates, and d is the separation between the plates.

Using the given values:

C = (8.85 × 10-12 F/m)(9.92 cm2)/(3.20 mm)

Simplifying the units:

C = (8.85 × 10-12 F/m)(9.92 × 10-4 m2)/(3.20 × 10-3 m)

C = 27.45 × 10-12 F

The capacitance of the parallel-plate capacitor is approximately 27.45 picofarads (pF).