Let's solve the problem step by step in English, with explanations for each step. --- ## **Given Data** - Thicknesses: \( a_1 = 3\, \text{cm} \) \( a_2 = 3\, \text{cm} \) \( a_3 = 6\, \text{cm} \) - Relative permittivities: \( \varepsilon_{r1} = 4 \) \( \varepsilon_{r2} = 2 \) \( \varepsilon_{r3} = 6 \) - Breakdown fields: \( E_{d1} = 60\, \text{kV/cm} \) \( E_{d2} = 40\, \text{kV/cm} \) \( E_{d3} = 50\, \text{kV/cm} \) - Total voltage: \( U = 300\, \text{kV} \) --- ## **a) Find the Electric Field Intensity in Each Layer** ### **Step 1:** Use the boundary condition for electric displacement (\( D \)): The system is a series of dielectrics, so the electric displacement \( D \) is the same in each layer: \[ D = \varepsilon_0 \varepsilon_{r1} E_1 = \varepsilon_0 \varepsilon_{r2} E_2 = \varepsilon_0 \varepsilon_{r3} E_3 \] ### **Step 2:** Express voltages across each layer and sum to total voltage: \[ U = V_1 + V_2 + V_3 = E_1 a_1 + E_2 a_2 + E_3 a_3 \] But from Step 1: \[ E_1 = \frac{D}{\varepsilon_0 \varepsilon_{r1}}, \quad E_2 = \frac{D}{\varepsilon_0 \varepsilon_{r2}}, \quad E_3 = \frac{D}{\varepsilon_0 \varepsilon_{r3}} \] So, \[ U = \frac{D}{\varepsilon_0} \left( \frac{a_1}{\varepsilon_{r1}} + \frac{a_2}{\varepsilon_{r2}} + \frac{a_3}{\varepsilon_{r3}} \right) \] ### **Step 3:** Solve for \( D \): Plug in the values: \[ U = 300\, \text{kV} = 300,000\, \text{V} \] \[ a_1 = a_2 = 3\, \text{cm} = 0.03\, \text{m}, \quad a_3 = 6\, \text{cm} = 0.06\, \text{m} \] \[ \frac{a_1}{\varepsilon_{r1}} = \frac{0.03}{4} = 0.0075\, \text{m} \] \[ \frac{a_2}{\varepsilon_{r2}} = \frac{0.03}{2} = 0.015\, \text{m} \] \[ \frac{a_3}{\varepsilon_{r3}} = \frac{0.06}{6} = 0.01\, \text{m} \] Sum: \[ \left( \frac{a_1}{\varepsilon_{r1}} + \frac{a_2}{\varepsilon_{r2}} + \frac{a_3}{\varepsilon_{r3}} \right) = 0.0075 + 0.015 + 0.01 = 0.0325\, \text{m} \] \[ U = \frac{D}{\varepsilon_0} \times 0.0325 \] \[ D = \frac{U \varepsilon_0}{0.0325} \] \[ \varepsilon_0 = 8.854 \times 10^{-12}\, \text{F/m} \] \[ D = \frac{300,000 \times 8.854 \times 10^{-12}}{0.0325} = 8.18 \times 10^{-7}\, \text{C/m}^2 \] ### **Step 4:** Now calculate the field in each layer: \[ E_1 = \frac{D}{\varepsilon_0 \varepsilon_{r1}} = \frac{8.18 \times 10^{-7}}{8.854 \times 10^{-12} \times 4} \approx 23,100\, \text{V/m} = 2.31\, \text{kV/cm} \] \[ E_2 = \frac{D}{\varepsilon_0 \varepsilon_{r2}} = \frac{8.18 \times 10^{-7}}{8.854 \times 10^{-12} \times 2} \approx 46,200\, \text{V/m} = 4.62\, \text{kV/cm} \] \[ E_3 = \frac{D}{\varepsilon_0 \varepsilon_{r3}} = \frac{8.18 \times 10^{-7}}{8.854 \times 10^{-12} \times 6} \approx 15,400\, \text{V/m} = 1.54\, \text{kV/cm} \] ### **Explanation:** - We used the fact that the displacement field \( D \) is constant across all layers. - The voltage divides according to the permittivity and thickness of each dielectric. --- ## **b) Will Any Layer Be Punctured?** Compare the calculated electric field in each layer with its breakdown field: - \( E_1 = 2.31\, \text{kV/cm} < E_{d1} = 60\, \text{kV/cm} \) - \( E_2 = 4.62\, \text{kV/cm} < E_{d2} = 40\, \text{kV/cm} \) - \( E_3 = 1.54\, \text{kV/cm} < E_{d3} = 50\, \text{kV/cm} \) **No layer will be punctured at 300 kV.** --- **If you want to proceed to part c), let me know!**

Certainly! Here's a detailed explanation in words, broken down into two steps: --- ### **Step 1: Find the Electric Field Intensity in Each Layer** **Explanation:** In a multilayer dielectric system, the electric displacement \( D \) remains constant across all layers because there are no free charges within the dielectric materials. This means the amount of electric flux passing through each layer is the same. Since \( D = \varepsilon_0 \varepsilon_{r} E \), where \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_{r} \) is the relative permittivity of the material, and \( E \) is the electric field, we can relate the fields in each layer through \( D \). First, we calculate the total voltage applied across the entire stack. The voltage divides among the layers based on their thicknesses and dielectric properties. By summing the voltage drops across each layer, which depend on their electric fields and thicknesses, and using the relation for \( D \), we solve for the electric field in each layer. The calculations show that the electric field in each layer is much below the breakdown field, meaning no layer is at risk of puncture at this voltage. --- ### **Step 2: Determine if Any Layer Will Puncture** **Explanation:** To assess whether any layer will puncture (break down), compare the calculated electric field in each layer with its known breakdown field (the maximum field the material can withstand without failure). Since the electric fields in all layers are significantly below their respective breakdown fields, none of the layers will puncture at the applied voltage of 300 kV. This indicates that the system is safe under these conditions, and the dielectric materials can withstand this voltage without failure. --- **Summary:** - The electric fields in all layers are well within safe limits. - Therefore, the multilayer dielectric system can handle the applied voltage without any layer puncturing.