The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K_¹ and K_² with thickness \(\frac{3}{8}d\) and \(\frac{d}{2}\), 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:

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Question

This question was previously asked in

NEET 2025 Official Paper (Held On: 04 May, 2025)

Answer 1

Calculation:
1 (3)

C_eq = (ε₀ A) / (t₁/K₁ + t₂/K₂ + t₃/K₃)

Here C₀ = ε₀ A / d, t₁ = 3d / 8, t₂ = d / 2, t₃ = d / 8

K₁ = K₁, K₂ = K₁ / 1.25, K₃ = 1

Given C_eq = 2C₀

⇒ 2C₀ = ε₀ A / ( (3d / 8K₁) + (d × 1.25 / 2K₁) + (d / 8) )

⇒ 2ε₀ A / d = ε₀ A / ( (3d / 8K₁) + (d / 2K₁) + (d / 8) )

⇒ 2 = 1 / ( (3 / 8K₁) + (5 / 8K₁) + (1 / 8) )

⇒ K₁ = 8 / 3 = 2.66