Question

The plates of a parallel-plate capacitor are 3.26 mm apart, and each has an area of 9.72 cm^2. Each plate carries a charge of magnitude 4.60×10^(-8) C. The plates are in a vacuum. What is the capacitance of this capacitor? A. 141 pF B. 67.89 pF C. 2.44 nF D. 4.60 pF

Answer 1

The capacitance of the parallel-plate capacitor can be calculated using the formula C = (ε₀ * A) / d, where ε₀ is the vacuum permittivity, A is the area of each plate, and d is the separation between the plates. Plugging in the values given in the question, the capacitance is found to be 2.44 nF.

Step-by-step explanation:

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

C = (ε₀ × A) / d

Where C is the capacitance, ε₀ is the vacuum permittivity (8.85 x 10^-12 F/m), A is the area of each plate, and d is the separation between the plates.

For this particular capacitor, with an area of 9.72 cm² and a separation of 3.26 mm, we can convert the units:

C = (8.85 x 10^-12 F/m × 0.0972 m²) / 0.00326 m

C = 2.44 x 10^-9 F

Therefore, the capacitance of this capacitor is 2.44 nF (option C).