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Question
JEE Main
Physics

Three parallel plate capacitors each with area $A$ and separation $d$ are filled with two dielectric ($k_1$ and $k_2$) in the following fashion. Which of the following is true? ($k_1 > k_2$)

The correct answer is
$C_B > C_C > C_A$

To solve this problem, let's analyze each capacitor configuration and calculate their capacitance.

Each capacitor has plate area \(A\) and separation \(d\). The capacitance of a parallel plate capacitor with a dielectric is given by:

\(C = \frac{{k \varepsilon_0 A}}{d}\)

  1. **Capacitor A:**
    • The dielectric layers are in series.
      • Capacitance with dielectric \(k_1\): \(C_1 = \frac{{k_1 \varepsilon_0 A}}{d/2}\)
      • Capacitance with dielectric \(k_2\): \(C_2 = \frac{{k_2 \varepsilon_0 A}}{d/2}\)
    • For capacitors in series: \(\frac{1}{C_A} = \frac{1}{C_1} + \frac{1}{C_2}\)
  2. **Capacitor B:**
    • The dielectric layers are still in series but with different configuration.
      • Capacitance with \(k_2\) (top half): \(C_1 = \frac{{k_2 \varepsilon_0 A}}{d/2}\)
      • Capacitance with \(k_1\) (bottom half): \(C_2 = \frac{{k_1 \varepsilon_0 A}}{d/2}\)
    • For capacitors in series: \(\frac{1}{C_B} = \frac{1}{C_1} + \frac{1}{C_2}\)
  3. **Capacitor C:**
    • The dielectric layers are in parallel.
      • Capacitance with \(k_1\): \(C_1 = \frac{{k_1 \varepsilon_0 A/2}}{d}\)
      • Capacitance with \(k_2\): \(C_2 = \frac{{k_2 \varepsilon_0 A/2}}{d}\)
    • For capacitors in parallel: \(C_C = C_1 + C_2\)

Since \(k_1 > k_2\), the capacitor A and B calculations in series will differ in sums due to different \(k\) order in configuration, resulting in \(C_B > C_C > C_A\).

Thus, the correct answer is \(C_B > C_C > C_A\).

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