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The electric field is a crucial concept in understanding how parallel-plate capacitors work. It is essentially a measure of the force per unit charge exerted between the plates. In this context, the electric field ( E ) is uniform between the two plates of the capacitor, meaning its strength and direction are constant throughout the space between the plates. This uniform electric field is what allows capacitors to store energy.* Mathematically, the electric field in a capacitor is given by the equation: \[ E = \frac{V}{d} \] where \( V \) is the potential difference between the plates, and \( d \) is the separation distance between them. For the given problem, with an electric field of \( 4.00 \times 10^6 \ \mathrm{V/m} \), understanding this behavior is key to performing further calculations regarding potential difference and capacitor behavior.

The potential difference, also known as voltage, between the plates of a parallel-plate capacitor is the key to its ability to store energy. It is defined as the work done to move a unit charge between the plates against the electric field. The potential difference ( V ) can be calculated using the formula:\[ V = E \times d \]where \( E \) is the electric field strength, and \( d \) is the distance between the plates.* In this exercise, given \( E = 4.00 \times 10^6 \ \mathrm{V/m} \) and \( d = 2.50 \ \mathrm{mm} = 2.50 \times 10^{-3} \ \mathrm{m} \), the potential difference is calculated as \( 10,000 \ \mathrm{V} \). This high potential difference indicates a significant energy storage capability between the plates, enabling the capacitor to perform its functions effectively.

Capacitance is a measure of a capacitor's ability to store charge per unit voltage across its plates. This key electrical property directly affects how much charge a capacitor can hold at a given potential difference. The formula to calculate the capacitance \( C \) of a parallel-plate capacitor is:\[ C = \varepsilon_0 \frac{A}{d} \]where:

• \( \varepsilon_0 \) is the vacuum permittivity, a constant with a value of \( 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)} \).
• \( A \) is the plate area.
• \( d \) is the distance between the plates.

In the problem, using the previously determined values for the plate area and the distance, the capacitance was calculated to be approximately \( 8.01 \times 10^{-12} \ \mathrm{F} \), or Farads, which is typical for capacitors used in electronic circuits.

The area of the plates in a parallel-plate capacitor is crucial as it directly influences the capacitance. Larger plate areas allow more charge storage by increasing the capacitance. To determine the plate area \( A \), we use the relation involving charge \( Q \), electric field \( E \), and vacuum permittivity \( \varepsilon_0 \):\[ E = \frac{Q}{\varepsilon_0 \times A} \]Reorganizing this equation allows us to solve for plate area:\[ A = \frac{Q}{E \times \varepsilon_0} \]From the given data, with a charge of \( 80.0 \ \mathrm{nC} \) (or \( 80.0 \times 10^{-9} \ \mathrm{C} \)) and \( E = 4.00 \times 10^6 \ \mathrm{V/m} \), the area comes out to approximately \( 2.26 \times 10^{-3} \ \mathrm{m^2} \). This value of plate area contributes to the overall efficiency and capacity of the capacitor.

Vacuum permittivity, denoted as \( \varepsilon_0 \), is a fundamental constant in electromagnetism. It characterizes the ability of the vacuum to permit electric field lines. This constant is vital in the formula calculating capacitance and appears in equations governing electric fields in capacitors.* The value of vacuum permittivity is \( 8.85 \times 10^{-12} \ \mathrm{C^2/(N \cdot m^2)} \). This constant helps determine how effective a capacitor is at storing energy between its plates in vacuum. It reflects the relationship between electric force and electric field lines in a non-material medium, ultimately contributing to predicting how capacitors will behave in electrical circuits.