Three parallel plate capacitors each with area A and separation d are filled with two dielectric ( k₁ and k₂ ) in the following fashion. Which of the following is true?
(k₁>k₂)

Solution

Physics sir · 3 steps

Split each capacitor into blocks, then join them in series or parallel the way the field sees them.

1. Building blocks. Let C₀ = (ε₀ A)/d. A block of half the area and half the gap gives (kε₀ (A/2))/(d/2) = kC₀. A full-width slab of thickness d/2 gives 2kC₀. Blocks stacked between the plates are in series, and blocks side by side are in parallel.

   Capacitor	How the blocks combine	Capacitance / C₀	With k₁ = 2, k₂ = 1
   A	2k₁ in series with (k₁ + k₂)	2k₁(k₁+k₂)/(3k₁+k₂)	12/7 ≈ 1.71
   B	2k₂ in series with (k₁ + k₂)	2k₂(k₁+k₂)/(k₁+3k₂)	6/5 = 1.2
   C	two columns in parallel, each k₁ in series with k₂	2k₁k₂/(k₁+k₂)	4/3 ≈ 1.33

2. Check for any k. C_A/C_C = (k₁+k₂)²/k₂(3k₁+k₂), and (k₁+k₂)² - k₂(3k₁+k₂) = k₁(k₁-k₂) > 0, so C_A > C_C. Likewise, C_C/C_B = k₁(k₁+3k₂)/(k₁+k₂)², and k₁(k₁+3k₂) - (k₁+k₂)² = k₂(k₁-k₂) > 0, so C_C > C_B.
3. So. C_A > C_C > C_B, which is option (c).

The idea to keep. Layers stacked between the plates add like resistors in parallel (series capacitors). Strips side by side add directly. Draw the block circuit first, then compare with easy numbers like k₁ = 2, k₂ = 1.

Answer from the NTA final answer key. Solution by PrepWiser.