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An air-filled parallel-plate capacitor has plates of area 3.00 cm² separated by 2.40 mm. The capacitor is connected to a(n) 13.0 V battery. Find the value of its capacitance. pF

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First, let's express the area and the distance in the International System of Units (SI). The area is 3.00 cm², which we know is equivalent to 3.00 × 10⁻⁴ m² because 1 m² = 10,000 cm². The distance between the plates is 2.40 mm, which is 2.40 × 10⁻³ m, since 1 m = 1000 mm.

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

C = ε₀ × (A/d)

Where:
C is the capacitance,
ε₀ is the permittivity of free space,
A is the area of one of the capacitor's plates,
d is the distance between the plates.

We know that the permittivity of free space (ε₀) is 8.85 × 10⁻¹² F/m.

So, substituting the values into the formula gives us:

C = (8.85 × 10⁻¹² F/m) × [(3.00 × 10⁻⁴ m²) / (2.40 × 10⁻³ m)]

The units of meters (m) and square meters (m²) cancel out, leaving us with a capacitance in Farads (F).

However, in this context, it's more convenient to express our answer in picoFarads (pF), since 1 Farad is a very large unit of capacitance. We can convert from Farads to picoFarads by multiplying our result by 10¹² (since 1 F = 10¹² pF).

Doing the calculation and conversion, we find that the capacitance of the capacitor is approximately 1.10625 pF.

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