163k views
4 votes
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

1 Answer

3 votes

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.

User Ang Lee
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.