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To understand how we calculate the capacitance of a parallel-plate capacitor, it's important to recognize its basic components: two plates separated by a certain distance. The capacitance is determined by the formula \( C = \frac{\varepsilon_0 A}{d} \). Here, \( \varepsilon_0 \), known as the permittivity of free space, is a constant valued at 8.85 \( \times 10^{-12} \ \text{F/m} \). This constant reflects how much electric field can seep through a vacuum, and it's universally used in these calculations.
The area \( A \) represents the surface area of one of the plates, which needs conversion from square centimeters to square meters using the conversion factor \( 1 \ \text{m}^2 = 10^{4} \ \text{cm}^2 \). For this problem, \( A = 9.82 \ \text{cm}^2 \) becomes \( 9.82 \times 10^{-4} \ \text{m}^2 \).
The distance \( d \) between the plates must also be converted from millimeters to meters: \( d = 3.28 \ \text{mm} = 3.28 \times 10^{-3} \ \text{m} \).

• This conversion is necessary because SI units are essential for ensuring consistency in physical calculations.
• After inputting these values into the capacitance formula, you find \( C \approx 2.65 \times 10^{-12} \ \text{F} \), indicating a very small amount of capacitance due to the microscopic separation in vacuum-bound plates.

The concept of the electric field (\( E \)) in a parallel-plate capacitor is crucial for understanding how charge interacts over a distance. The electric field represents the force that a charge experiences per unit charge and is particularly straightforward in a uniform field context like between parallel plates.
In this scenario, the electric field's magnitude is determined using the potential difference between the plates and the separation distance, given by the relationship \( E = \frac{V}{d} \).

• This formula indicates that the electric field strength depends directly on the potential difference \( V \), which is the energy difference causing the charges to move.
• It inversely depends on \( d \), the separation between the plates, helping us understand how spacing affects field strength.

With a potential difference derived earlier as \( 1.64 \times 10^{4} \ \text{V} \) and \( d = 3.28 \times 10^{-3} \ \text{m} \), substitution into the formula provides an electric field \( E \approx 5.00 \times 10^{6} \ \text{V/m} \). This large value reflects the strong force exerted between closely spaced plates in a capacitor.

The potential difference, sometimes called voltage, across the plates of a parallel-plate capacitor is a measure of how much potential energy is distributed per unit charge between the plates. It can be calculated using the formula \( V = \frac{Q}{C} \), where \( Q \) is the charge on one of the plates and \( C \) is the capacitance.
In this context, \( Q = 4.35 \times 10^{-8} \ \text{C} \), illustrating a very small amount of charge, as is common in such microscopic systems.

• The capacitance, calculated as \( 2.65 \times 10^{-12} \ \text{F} \), is crucial to computing the potential difference.
• This formula inherently speaks to how the stored charge and spacing in the capacitor contribute to this energetic measure of potential difference.

Substituting the known values, you determine \( V \approx 1.64 \times 10^{4} \ \text{V} \), signifying a significant potential energy per charge unit. This energy availability is pivotal for capacitors' role in storing and discharging electricity in circuits.