A parallel-plate capacitor has plate gap d. Two dielectric slabs of constants K₁ and K₂, with thicknesses 3d/8 and d/2, are inserted; the remaining gap is air. The capacitance doubles. Given K₁ = 1.25K₂, find K₁. Image
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Question

A parallel-plate capacitor has plate gap d. Two dielectric slabs of constants K₁ and K₂, with thicknesses 3d/8 and d/2, are inserted; the remaining gap is air. The capacitance doubles. Given K₁ = 1.25K₂, find K₁. Image

Options

G1 Physics · Accepted Answer

Step 1: The uncovered air gap is d − 3d/8 − d/2 = d/8. Step 2: For series dielectric layers, C = ε₀A / [(3d/8)/K₁ + (d/2)/K₂ + d/8]. Step 3: Relative to C₀ = ε₀A/d, C/C₀ = 1/[3/(8K₁) + 1/(2K₂) + 1/8]. Step 4: K₁ = 1.25K₂ = (5/4)K₂, hence 1/K₂ = 5/(4K₁). Step 5: Effective denominator = 3/(8K₁) + 5/(8K₁) + 1/8 = 1/K₁ + 1/8. Step 6: Since C = 2C₀, 1/K₁ + 1/8 = 1/2. Thus 1/K₁ = 3/8 and K₁ = 8/3 ≈ 2.67. Final answer: 2.66 (nearest listed), Option A.