Q1.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:
- A2.66Answer
- B2.33
- C1.60
- D1.33
Slabs stacked along the field direction behave as capacitors in series, so C = ε₀A / Σ(tᵢ/Kᵢ), and note the leftover air gap of thickness d/8. Setting this equal to 2ε₀A/d and substituting K₂ = 0.8K₁ gives 1/K₁ = 3/8, so K₁ = 8/3 ≈ 2.66.
Source: NEET 2025 (4 May), Code 45, Q20