Space between the plates of a parallel plate capacitor of plate area 4 cm² and separation of (d) 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is ________ pF. (Given ε₀ = 8.85 × 10⁻¹² F/m)

 

🧠 Effective Capacitance of a Parallel Plate Capacitor with Dielectric Slabs 

📝 Problem Statement:

A parallel plate capacitor has plate area = 4 cm² and separation between plates = 1.77 mm. The space between the plates is filled with two uniform dielectric materials with dielectric constants 3 and 5, arranged side by side (i.e., in parallel configuration within the same capacitor). Another capacitor of capacitance 7.5 pF is connected in parallel to this setup.
Find the total effective capacitance of the system.
Given:

• ε₀ = 8.85 × 10⁻¹² F/m

• 1 cm² = 1×10⁻⁴ m²

• 1.77 mm = 1.77×10⁻³ m

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⚙️ Step-by-Step Solution:

✅ Step 1: Understanding the Geometry and Setup

Since the two dielectrics are placed side-by-side, the same distance (d) applies for both, but area is split equally (each slab gets half the area).

• Total area A = 4 cm² = 4 × 10⁻⁴ m²

• Area of each section = A/2 = 2 × 10⁻⁴ m²

• Distance d = 1.77 mm = 1.77 × 10⁻³ m

• Dielectric constants: k₁ = 3, k₂ = 5

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🧮 Step 2: Capacitance of Each Dielectric Part

We treat the dielectric-filled capacitor as two capacitors in parallel:

C₁ = (ε₀ × A₁ × k₁) / d
C₂ = (ε₀ × A₂ × k₂) / d

So,

C₁ = (8.85 × 10⁻¹²) × (2 × 10⁻⁴) × 3 / (1.77 × 10⁻³)  
    ≈ 3 × 8.85 × 2 / 1.77 × 10⁻¹³  
    = 53.1 / 1.77 × 10⁻¹³  
    ≈ 30 × 10⁻¹³ F = 3.0 pF

C₂ = (8.85 × 10⁻¹²) × (2 × 10⁻⁴) × 5 / (1.77 × 10⁻³)  
    ≈ 5 × 8.85 × 2 / 1.77 × 10⁻¹³  
    = 88.5 / 1.77 × 10⁻¹³  
    ≈ 50 × 10⁻¹³ F = 5.0 pF

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➕ Step 3: Add the Two Capacitances

Since they are in parallel:
C_parallel = C₁ + C₂ = 3.0 pF + 5.0 pF = 8.0 pF

Now, add the other capacitor of 7.5 pF (also in parallel):

C_total = 8.0 pF + 7.5 pF = 15.5 pF

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✅ Final Answer:

Effective Capacitance = 15.5 pF

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📌 Key Concept Recap:

• When dielectrics are placed side by side, treat the setup as parallel capacitors.

• Use C = (ε₀ × A × k) / d for each part.

• Add all parallel capacitances directly.

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🎯 JEE Tip:

In JEE questions with multiple dielectrics, always identify:

• Whether they are placed side by side (parallel)

• Or one over the other (series)
  Your interpretation of geometry changes the entire approach!