Question

the plates of a parallel-plate capacitor are 3.22 mm apart, and each has an area of 9.92 cm² . each plate carries a charge of magnitude 4.45×10⁻⁸ c . the plates are in vacuum. What is the capacitance? Express your answer with the appropriate units

Answer 1

Using the formula for the capacitance of a parallel-plate capacitor, the given dimensions result in a capacitance of 2.72 pF (picofarads).

Step-by-step explanation:

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

C = ε₀ * (A / d)

where ε₀ is the vacuum permittivity constant (ε₀ = 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 each plate of the capacitor has an area of 9.92 cm² (which is 9.92 × 10⁻´ m²) and the separation is 3.22 mm (which is 3.22 × 10⁻³ m), the capacitance can be calculated as follows:

C = 8.85 × 10⁻¹² F/m * (9.92 × 10⁻´ m² / 3.22 × 10⁻³ m)

C = 2.72 × 10⁻č F or 2.72 pF (picofarads)

Thus, the capacitance of the capacitor is 2.72 pF.