A parallel plate capacitor has plates of area A = 5.50 x 10-² m² separated by distance d = 2.37 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. 2.05e-9 ✓ F What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant x = 2.30 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) 2.85e-9 (c) figure (b) 1.90e-16 ✓ F X F A K b A 4

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A parallel plate capacitor has plates of area \( A = 5.50 \times 10^{-2} \, \text{m}^2 \) separated by a distance \( d = 2.37 \times 10^{-4} \, \text{m} \). The permittivity of free space is \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \).

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

- **Answer:** \( 2.05 \times 10^{-9} \, \text{F} \) ✔️

**What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant \( \kappa = 2.30 \) 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):** \( 2.85 \times 10^{-9} \, \text{F} \)
  
- **(c) Figure (b):** \( 1.90 \times 10^{-16} \, \text{F} \) ❌

### Explanation of Figures

Figures (a) and (b) illustrate two scenarios with a parallel plate capacitor:

- **Figure (a):** Shows the capacitor with half of the region filled with a dielectric material having a dielectric constant \( \kappa = 2.30 \) and the other half filled with air. This setup leads to a parallel combination of capacitors.

- **Figure (b):** Displays the capacitor with a similar division between air and dielectric but arranged in a manner that leads to a series combination of capacitors. 

The gray plates labeled \( A \) are the conducting plates, and the area between them is sectioned with the dielectric \( \kappa \) in yellow. The separation distance \( d \) is indicated between the plates.
Transcribed Image Text:A parallel plate capacitor has plates of area \( A = 5.50 \times 10^{-2} \, \text{m}^2 \) separated by a distance \( d = 2.37 \times 10^{-4} \, \text{m} \). The permittivity of free space is \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \). **(a)** Calculate the capacitance (in F) if the space between the plates is filled with air. - **Answer:** \( 2.05 \times 10^{-9} \, \text{F} \) ✔️ **What is the capacitance (in F) if the space is filled half with air and half with a dielectric of constant \( \kappa = 2.30 \) 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):** \( 2.85 \times 10^{-9} \, \text{F} \) - **(c) Figure (b):** \( 1.90 \times 10^{-16} \, \text{F} \) ❌ ### Explanation of Figures Figures (a) and (b) illustrate two scenarios with a parallel plate capacitor: - **Figure (a):** Shows the capacitor with half of the region filled with a dielectric material having a dielectric constant \( \kappa = 2.30 \) and the other half filled with air. This setup leads to a parallel combination of capacitors. - **Figure (b):** Displays the capacitor with a similar division between air and dielectric but arranged in a manner that leads to a series combination of capacitors. The gray plates labeled \( A \) are the conducting plates, and the area between them is sectioned with the dielectric \( \kappa \) in yellow. The separation distance \( d \) is indicated between the plates.
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