Question

The plates of a parallel-plate capacitor are 3.28 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 of the capacitor?

Answer 1

The capacitance of a parallel-plate capacitor with plates 3.28 mm apart and each having an area of 9.92 cm² in vacuum is 2.672 picofarads (pF).

Step-by-step explanation:

To calculate the capacitance of a parallel-plate capacitor in vacuum, we can use the formula:

C = ε_0 * (A / d)

where C is the capacitance, ε_0 is the vacuum permittivity (8.854 x 10-12 F/m), A is the area of one plate in square meters, and d is the distance between the plates in meters.

First, we convert the area from cm² to m² and the distance from mm to m:

• Area (A) = 9.92 cm² = 9.92 x 10-4 m²
• Distance (d) = 3.28 mm = 3.28 x 10-3 m

Now we can substitute these values into the formula:

C = (8.854 x 10-12 F/m) * (9.92 x 10-4 m² / 3.28 x 10-3 m)

Calculating this gives us the capacitance C:

C = 2.672 x 10-12 F = 2.672 pF

The capacitance of the capacitor is therefore 2.672 picofarads (pF).