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This problem involves a parallel-plate capacitor with three regions of different dielectric materials. The key to solving it is understanding how to treat the system of multiple dielectrics.

To find the total capacitance, we approach the system as a combination of capacitors in series, with each section having a different dielectric constant.

### Here's the step-by-step process to solve this:

1. **Given data:**
   - Plate area: \( A = 10.7 \, \text{cm}^2 = 10.7 \times 10^{-4} \, \text{m}^2 \)
   - Plate separation: \( 2d = 6.91 \, \text{mm} \) so \( d = 3.455 \, \text{mm} = 3.455 \times 10^{-3} \, \text{m} \)
   - Dielectric constants: 
     - \( \kappa_1 = 21.4 \)
     - \( \kappa_2 = 44.4 \)
     - \( \kappa_3 = 56.8 \)

2. **Divide the capacitor into sections:**
   - The left half is filled with \( \kappa_1 \) and has a dielectric thickness of \( d \).
   - The right half is divided into two parts:
     - The top part is filled with \( \kappa_2 \), having a dielectric thickness of \( d \).
     - The bottom part is filled with \( \kappa_3 \), having a dielectric thickness of \( d \).

3. **Calculate the capacitance for each section:**

   The capacitance for a single region with dielectric is given by:
   \[
   C = \frac{\kappa \varepsilon_0 A}{d}
   \]
   Where:
   - \( \kappa \) is the dielectric constant.
   - \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space.
   - \( A \) is the area of the plates.
   - \( d \) is the separation between the plates.

4. **Apply this formula to each region:**
   - For the first region (left half with \( \kappa_1 \)): 
     \[
     C_1 = \frac{\kappa_1 \varepsilon_0 A}{d}
     \]
   - For the second region (top right half with \( \kappa_2 \)): 
     \[
     C_2 = \frac{\kappa_2 \varepsilon_0 A}{d}
     \]
   - For the third region (bottom right half with \( \kappa_3 \)): 
     \[
     C_3 = \frac{\kappa_3 \varepsilon_0 A}{d}
     \]

5. **Total capacitance:**
   The total capacitance \( C_{\text{total}} \) is the parallel combination of the individual capacitances, since they are aligned in parallel due to their shared plate area \( A \):
   \[
   C_{\text{total}} = C_1 + C_2 + C_3
   \]

6. **Plugging in the values:**
   \[
   C_{\text{total}} = \frac{\varepsilon_0 A}{d} \left( \kappa_1 + \kappa_2 + \kappa_3 \right)
   \]
   Now, substitute the known values:
   \[
   C_{\text{total}} = \frac{(8.85 \times 10^{-12}) (10.7 \times 10^{-4})}{3.455 \times 10^{-3}} (21.4 + 44.4 + 56.8)
   \]

   This will give you the final capacitance value.

Let me know if you'd like me to continue with the calculation or if you have any questions!

The figure shows a parallel-plate capacitor of plate area A = 10.7 cm² and plate separation 2d = 6.91 mm. The left half of the gap is filled with material of dielectric constant κ₁ = 21.4; the top of the right half is filled with material of dielectric constant κ₂ = 44.4; the bottom of the right half is filled with material of dielectric constant κ₃ = 56.8. What is the capacitance?

Learn how to calculate the capacitance of a parallel-plate capacitor with multiple dielectrics. This problem involves dielectrics with constants κ₁ = 21.4, κ₂ = 44.4, and κ₃ = 56.8, plate area 10.7 cm², and plate separation 6.91 mm. Step-by-step solution included.

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