91Ó°ÊÓ

To calculate the capacitance of a parallel-plate capacitor, you need to know the formula: \( C = \frac{\varepsilon_0 \cdot A}{d} \), where:

• \( C \) is the capacitance in farads (F).
• \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).
• \( A \) is the area of one plate in square meters.
• \( d \) is the separation distance between the plates in meters.

For our specific problem, the area \( A = (0.16 \, \text{m})^2 = 0.0256 \, \text{m}^2 \) and the distance \( d = 0.0047 \, \text{m} \). Substituting these values into the formula gives us \( C = \frac{8.85 \times 10^{-12} \times 0.0256}{0.0047} = 4.81 \times 10^{-12} \, \text{F} \).
Understanding this calculation helps us realize how changes in the plate area or separation affect capacitance. If the plates are closer or have a larger area, they can store more charge. Conversely, increasing the distance reduces capacitance.

The electric field \( E \) between the plates of a capacitor can be found using the formula: \( E = \frac{V}{d} \), where:

• \( E \) is the electric field in volts per meter (V/m).
• \( V \) is the voltage across the plates.
• \( d \) is the spacing between the plates in meters.

For the given capacitor, we use \( V = 12 \, \text{V} \) and \( d = 0.0047 \, \text{m} \) to find that \( E = \frac{12}{0.0047} = 2553.19 \, \text{V/m} \).
This electric field measures the force felt by a charge placed between the plates. The larger the field, the stronger the force experienced by charges. It's crucial for understanding how capacitors control electric forces in circuits.
When the separation increases to \( d = 0.0094 \, \text{m} \), the field becomes \( 2588.38 \, \text{V/m} \), reflecting reduced influence due to increased separation when charge remains constant.

In capacitors, the stored energy \( U \) is calculated with the formula: \( U = \frac{1}{2} C V^2 \), where:

• \( U \) is the energy in joules (J).
• \( C \) is the capacitance in farads.
• \( V \) is the voltage across the capacitor.

For the initial configuration, substituting \( C = 4.81 \times 10^{-12} \, \text{F} \) and \( V = 12 \, \text{V} \) gives: \( U = \frac{1}{2} \times 4.81 \times 10^{-12} \times 12^2 = 3.46 \times 10^{-10} \, \text{J} \).
This energy is the work done to charge the capacitor. If we disconnect the battery and increase the distance to \( d = 0.0094 \, \text{m} \), the capacitance changes, as does the stored energy to \( 6.91 \times 10^{-10} \, \text{J} \), showing that when plates are separated with constant charge, energy stored increases.
This concept is important for understanding energy conservation and transfer in circuits involving capacitors.

Charge displacement in a capacitor refers to how charges are stored or moved between the plates. The charge \( Q \) on the capacitor plates can be found using: \( Q = C \cdot V \), where:

• \( Q \) is the charge in coulombs (C).
• \( C \) is the capacitance in farads.
• \( V \) is the potential difference across the plates.

Initially, \( Q = 4.81 \times 10^{-12} \cdot 12 = 5.77 \times 10^{-11} \, \text{C} \), illustrating that the charge is directly proportional to both capacitance and voltage.
When the battery is disconnected and the plate separation increases, the capacitance changes but the charge remains constant at \( 5.77 \times 10^{-11} \, \text{C} \) because no charge flows once the battery is removed.
This demonstrates that the source (battery) affects how and when charges are moved. It's crucial for scenarios where a constant charge is maintained, important for stable circuit operations.