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Capacitance is a measure of a capacitor's ability to store charge. For a parallel-plate capacitor, this can be calculated using the formula \[ C = \varepsilon_0 \frac{A}{d} \]where

• \( C \) is the capacitance in farads (F),
• \( \varepsilon_0 = 8.854 \times 10^{-12} \, \mathrm{F/m} \) is the permittivity of free space,
• \( A \) is the area of one plate in square meters, and
• \( d \) is the distance between the plates in meters.

To find the capacitance for a capacitor with an area of \( 40 \, \mathrm{cm}^2 \) and a separation of \( 1.0 \, \mathrm{mm} \), these values need to be converted to meters to match the units of \( \varepsilon_0 \). The area \( A \) becomes \( 4.0 \times 10^{-3} \, \mathrm{m}^2 \) and the distance \( d \) becomes \( 1.0 \times 10^{-3} \, \mathrm{m} \). Substituting into the formula gives:\[ C = 8.854 \times 10^{-12} \, \mathrm{F/m} \times \frac{4.0 \times 10^{-3} \, \mathrm{m}^2}{1.0 \times 10^{-3} \, \mathrm{m}} = 35.4 \, \mathrm{pF}. \] This is the capacitance of the given parallel-plate capacitor.

The electric field within a parallel-plate capacitor can be determined using the potential difference between the plates and the separation distance. The formula for the electric field \( E \) is:\[ E = \frac{V}{d} \]where:

• \( V \) is the potential difference across the plates in volts (V), and
• \( d \) is the spacing between the plates in meters.

For a capacitor charged to \( 500 \, \mathrm{V} \) with plates \( 1.0 \, \mathrm{mm} \) apart, this equation becomes:\[ E = \frac{500 \, \mathrm{V}}{1.0 \times 10^{-3} \, \mathrm{m}} = 5.0 \times 10^{5} \, \mathrm{V/m}. \]This high electric field is due to the close proximity of the plates, which concentrates the electric force between them.

The energy stored in a capacitor is an essential concept when considering its application in circuits. The amount of energy \( U \) stored can be found with:\[ U = \frac{1}{2} C V^2 \]where:

• \( U \) is the stored energy in joules (J),
• \( C \) is the capacitance in farads (F), and
• \( V \) is the potential difference in volts (V).

For the given capacitor, the values \( 35.4 \, \mathrm{pF} \) and \( 500 \, \mathrm{V} \) yield:\[ U = \frac{1}{2} \times 35.4 \times 10^{-12} \, \mathrm{F} \times (500 \, \mathrm{V})^2 = 4.425 \times 10^{-6} \, \mathrm{J} = 4.425 \, \mathrm{\mu J}. \] This shows that even a small capacitance can store a significant amount of energy when charged to a high voltage.

The charge \( Q \) on each plate of a capacitor is directly proportional to both the capacitance and the voltage across the plates. The relationship is expressed as:\[ Q = C \times V \]Here,

• \( Q \) is the charge in coulombs (C),
• \( C \) is the capacitance in farads (F), and
• \( V \) is the potential difference in volts (V).

For the parallel-plate capacitor in the exercise, with \( C = 35.4 \, \mathrm{pF} \) and \( V = 500 \, \mathrm{V} \), the charge is calculated as:\[ Q = 35.4 \times 10^{-12} \, \mathrm{F} \times 500 \, \mathrm{V} = 17.7 \times 10^{-9} \, \mathrm{C} = 17.7 \, \mathrm{nC}. \] This means that each plate holds a charge of \( 17.7 \, \mathrm{nC} \), illustrating the capacitor's ability to store electrical charge efficiently.