Question:

Please provide the below content same as it is in rendered equations form without any errors Sure — let’s **solve it again carefully from scratch**, very cleanly and **double-check every step**. I’ll keep it slow, logical, and **correct**. --- ## 1. Understanding the system (very important) * Parallel-plate capacitor * **Three dielectric layers in series (stacked vertically)** * Electric field is **perpendicular to layers** * Same **cross-sectional area** * Therefore → **series dielectric system** Top electrode: **300 kV** Bottom electrode: **ground** --- ## 2. Given data | Layer | Thickness (cm) | εr | Dielectric strength (kV/cm) | | ----- | -------------- | ------- | --------------------------- | | 1 | a₁ = 3 cm | εr₁ = 4 | Ed₁ = 60 | | 2 | a₂ = 3 cm | εr₂ = 2 | Ed₂ = 40 | | 3 | a₃ = 6 cm | εr₃ = 6 | Ed₃ = 30 | --- ## 3. Key physics rule (series dielectrics) ### Electric displacement **D is the same in all layers** [ D = \varepsilon_0 \varepsilon_r E ] So, [ E_i = \frac{D}{\varepsilon_0 \varepsilon_{ri}} ] --- ## 4. Total voltage equation [ V = E_1 a_1 + E_2 a_2 + E_3 a_3 ] Substitute (E_i): [ V = \frac{D}{\varepsilon_0} \left( \frac{a_1}{\varepsilon_{r1}} + \frac{a_2}{\varepsilon_{r2}} + \frac{a_3}{\varepsilon_{r3}} \right) ] --- ## 5. Compute the geometric–dielectric sum [ \frac{a_1}{\varepsilon_{r1}} = \frac{3}{4} = 0.75 ] [ \frac{a_2}{\varepsilon_{r2}} = \frac{3}{2} = 1.5 ] [ \frac{a_3}{\varepsilon_{r3}} = \frac{6}{6} = 1 ] [ \sum \frac{a_i}{\varepsilon_{ri}} = 0.75 + 1.5 + 1 = \boxed{3.25\ \text{cm}} ] --- ## 6. Case (a): Apply **300 kV** ### Step 1: Find ( \frac{D}{\varepsilon_0} ) [ \frac{D}{\varepsilon_0} ======================= # \frac{300}{3.25} \boxed{92.31\ \text{kV/cm}} ] --- ### Step 2: Electric field in each layer #### Layer 1 [ E_1 = \frac{92.31}{4} = \boxed{23.08\ \text{kV/cm}} ] #### Layer 2 [ E_2 = \frac{92.31}{2} = \boxed{46.15\ \text{kV/cm}} ] #### Layer 3 [ E_3 = \frac{92.31}{6} = \boxed{15.38\ \text{kV/cm}} ] --- ### Step 3: Compare with dielectric strengths | Layer | E (kV/cm) | Ed (kV/cm) | Result | | ----- | --------- | ---------- | ----------- | | 1 | 23.08 | 60 | Safe | | 2 | **46.15** | **40** | ❌ Breakdown | | 3 | 15.38 | 30 | Safe | --- ### ✅ **Answer (a)** ✔ **Yes, breakdown occurs** ✔ Breakdown happens **in Layer 2** --- ## 7. Case (b): Maximum voltage without breakdown The **weakest layer** limits the voltage → **Layer 2** Set: [ E_2 = E_{d2} = 40\ \text{kV/cm} ] --- ### Step 1: Express (E_2) in terms of voltage [ E_2 = \frac{1}{\varepsilon_{r2}} \cdot \frac{V_{\max}}{3.25} ] [ 40 = \frac{1}{2} \cdot \frac{V_{\max}}{3.25} ] --- ### Step 2: Solve for (V_{\max}) [ V_{\max} = 40 \times 2 \times 3.25 ] [ \boxed{V_{\max} = 260\ \text{kV}} ] --- ## 8. Final verified answers ✅ ### **(a)** The system **will break down at 300 kV**, and breakdown occurs in **Layer 2**. ### **(b)** The **maximum voltage without breakdown** is: [ \boxed{260\ \text{kV}} ] --- If you want, I can now: * Show **why low εr layers experience higher field** * Solve using **capacitor equivalent method** * Add a **short exam-ready explanation** * Draw a **field distribution diagram** Just tell me ?