A parallel plate capacitor has plates of area A = 8.00 x 10-2 m² separated by distance d = 2.22 x 10-4 m. (The permittivity of free space is = 8.85 x 10-12 c²/(N m²).) HINT (a) Calculate the capacitance (in F) if the space between the plates is filled with air. What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant x = 3.10 as in figure (a), and figure (b)? (Hint: One of the capacitors is a parallel combination and the other is a series combination.) (b) figure (a) (c) figure (b) a F A K b A Ⓡ

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**Parallel Plate Capacitor Analysis**

A parallel plate capacitor has plates of area \( A = 8.00 \times 10^{-2} \, \text{m}^2 \) separated by a distance \( d = 2.22 \times 10^{-4} \, \text{m} \). The permittivity of free space is given as \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \).

1. **Calculate the capacitance (in F) if the space between the plates is filled with air.**

   \[
   \text{Capacitance with air:} \quad \_\_\_\_ \, \text{F}
   \]

2. **Determine the capacitance (in F) if the space is half-filled with air and half with a dielectric of constant \( \kappa = 3.10 \), as shown in figures (a) and (b). Hint: One of the configurations is a parallel combination, while the other is a series combination.**

   - **Figure (a):**
     - The plates are divided horizontally, having both air and a dielectric on one plate.

   - **Figure (b):**
     - The plates are divided vertically, having air on one side and a dielectric on the other.

   \[
   \text{Figure (a):} \quad \_\_\_\_ \, \text{F}
   \]

   \[
   \text{Figure (b):} \quad \_\_\_\_ \, \text{F}
   \]

**Diagram Explanation:**

- **Figure (a)** displays a capacitor with a half-region filled with air and the other half filled with a dielectric material (constant \( \kappa = 3.10 \)). This scenario is expected to be treated as a parallel combination of capacitors given the configuration of layers.

- **Figure (b)** shows a capacitor where one half of the plates are in contact with air, while the other half with a dielectric, implying a series combination of the two segments along the line of separation of the materials.

This setup demonstrates the different approaches to solving composite dielectric arrangements in capacitor systems.
Transcribed Image Text:**Parallel Plate Capacitor Analysis** A parallel plate capacitor has plates of area \( A = 8.00 \times 10^{-2} \, \text{m}^2 \) separated by a distance \( d = 2.22 \times 10^{-4} \, \text{m} \). The permittivity of free space is given as \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \). 1. **Calculate the capacitance (in F) if the space between the plates is filled with air.** \[ \text{Capacitance with air:} \quad \_\_\_\_ \, \text{F} \] 2. **Determine the capacitance (in F) if the space is half-filled with air and half with a dielectric of constant \( \kappa = 3.10 \), as shown in figures (a) and (b). Hint: One of the configurations is a parallel combination, while the other is a series combination.** - **Figure (a):** - The plates are divided horizontally, having both air and a dielectric on one plate. - **Figure (b):** - The plates are divided vertically, having air on one side and a dielectric on the other. \[ \text{Figure (a):} \quad \_\_\_\_ \, \text{F} \] \[ \text{Figure (b):} \quad \_\_\_\_ \, \text{F} \] **Diagram Explanation:** - **Figure (a)** displays a capacitor with a half-region filled with air and the other half filled with a dielectric material (constant \( \kappa = 3.10 \)). This scenario is expected to be treated as a parallel combination of capacitors given the configuration of layers. - **Figure (b)** shows a capacitor where one half of the plates are in contact with air, while the other half with a dielectric, implying a series combination of the two segments along the line of separation of the materials. This setup demonstrates the different approaches to solving composite dielectric arrangements in capacitor systems.
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