The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric const

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Question

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K₁ and K₂ with thickness 3d/8 and d/2 respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K₁ = 1.25 K₂, the value of K₁ is:

Accepted Answer

Correct Answer: B — K₁ = 2.66Step-by-step Solution:The two dielectric slabs and the remaining air gap (thickness = d − 3d/8 − d/2 = d/8) act as three capacitors in series.Given: K₁ = 1.25K₂, Ceq = 2C₀, C₀ = ε₀A/dEquivalent capacitance (three dielectrics in series):1/Ceq = t₁/(K₁ε₀A) + t₂/(K₂ε₀A) + t₃/(ε₀A)where t₁ = 3d/8, t₂ = d/2, t₃ = d/8, K₃ = 1 (air)Since Ceq = 2C₀ = 2ε₀A/d:d/(2ε₀A) = (3d/8)/(K₁ε₀A) + (d/2)/(K₂ε₀A) + (d/8)/(ε₀A)1/2 = 3/(8K₁) + 1/(2K₂) + 1/8Using K₁ = 1.25K₂ = (5/4)K₂, so K₂ = 4K₁/5:1/2 − 1/8 = 3/(8K₁) + 5/(8K₁) = 8/(8K₁) = 1/K₁3/8 = 1/K₁ → K₁ = 8/3 ≈ 2.66Key Concepts from Electrostatic Potential and Capacitance (NCERT Class 12, Chapter 2):Capacitance of a parallel plate capacitor with dielectric: C = Kε₀A/d.When two dielectric slabs of different K fill different regions of separation, they act as capacitors in series (different thicknesses, same area): 1/Ctotal = t₁/(K₁ε₀A) + t₂/(K₂ε₀A) + ...The air gap also acts as a dielectric with K = 1.Dielectric constant K > 1 always. Inserting a dielectric increases capacitance (C' = KC) and decreases electric field inside (E' = E/K).

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