If you want to build the equivalent capacitor with a parallel plate capacitor with dielectric material K = 2.0. What is the size of the square plate, if the separation is the size of the plate? C₁ = 12 pF C₂ = 24 pF &₁ = 15 V C3 = 25 pF C5 = 16 pF C4 = 15 pF

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**Problem Statement:**

If you want to build the equivalent capacitor with a parallel plate capacitor with dielectric material κ = 2.0, what is the size of the square plate if the separation is ¼ the size of the plate?

**Given Circuit:**

A circuit diagram is provided which contains five capacitors interconnected in the following manner:

- Capacitor \( C_1 \) (12 pF) is in series with Capacitor \( C_5 \) (16 pF).
- This series combination is connected in parallel with Capacitor \( C_2 \) (24 pF).
- Capacitor \( C_3 \) (25 pF) is in series with the parallel combination of Capacitor \( C_4 \) (15 pF) and a branch containing Capacitors \( C_2 \) and the series combination of \( C_1 \) and \( C_5 \).
- The entire circuit is powered by an external voltage source \( ε_1 = 15V \).

**Explanation of the Circuit Diagram:**

1. **Series Connection:** Capacitors connected end-to-end, so the total capacitance \( C_{series} \) is:
   \[
   \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_5}
   \]

2. **Parallel Connection:** Capacitors connected across the same two points, so the total capacitance \( C_{parallel} \) is:
   \[
   C_{parallel} = C_2 + C_{series}
   \]

   Where \( C_{series} \) is the equivalent capacitance of \( C_1 \) and \( C_5 \). 

3. **Combined Connections:** The combined capacitance, including further series or parallel connections.

This setup aims to analyze the resultant capacitance and size the new parallel plate capacitor with the given dielectric and dimensions. The size of the plate \( A \) can be found using the formula:

\[ 
C = \kappa \epsilon_0 \left(\frac{A}{d}\right)
\]

Where:
- \( C \) is the total capacitance found.
- \( \kappa \) is the dielectric constant (given as 2.0).
- \( \epsilon_0 \) is the permittivity of free space (\( \approx 8.85 \times 10^{-
Transcribed Image Text:**Problem Statement:** If you want to build the equivalent capacitor with a parallel plate capacitor with dielectric material κ = 2.0, what is the size of the square plate if the separation is ¼ the size of the plate? **Given Circuit:** A circuit diagram is provided which contains five capacitors interconnected in the following manner: - Capacitor \( C_1 \) (12 pF) is in series with Capacitor \( C_5 \) (16 pF). - This series combination is connected in parallel with Capacitor \( C_2 \) (24 pF). - Capacitor \( C_3 \) (25 pF) is in series with the parallel combination of Capacitor \( C_4 \) (15 pF) and a branch containing Capacitors \( C_2 \) and the series combination of \( C_1 \) and \( C_5 \). - The entire circuit is powered by an external voltage source \( ε_1 = 15V \). **Explanation of the Circuit Diagram:** 1. **Series Connection:** Capacitors connected end-to-end, so the total capacitance \( C_{series} \) is: \[ \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_5} \] 2. **Parallel Connection:** Capacitors connected across the same two points, so the total capacitance \( C_{parallel} \) is: \[ C_{parallel} = C_2 + C_{series} \] Where \( C_{series} \) is the equivalent capacitance of \( C_1 \) and \( C_5 \). 3. **Combined Connections:** The combined capacitance, including further series or parallel connections. This setup aims to analyze the resultant capacitance and size the new parallel plate capacitor with the given dielectric and dimensions. The size of the plate \( A \) can be found using the formula: \[ C = \kappa \epsilon_0 \left(\frac{A}{d}\right) \] Where: - \( C \) is the total capacitance found. - \( \kappa \) is the dielectric constant (given as 2.0). - \( \epsilon_0 \) is the permittivity of free space (\( \approx 8.85 \times 10^{-
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