Question:

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. (\(k_1>k_2\)) Which of the following is true? 

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In mixed dielectric problems, identify whether dielectrics combine in series or parallel to compare capacitances quickly.
Updated On: Feb 5, 2026
  • \( C_B>C_C>C_A \)
  • \( C_C>C_A>C_B \)
  • \( C_C>C_B>C_A \)
  • \( C_A>C_C>C_B \)
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The Correct Option is A

Solution and Explanation

Step 1: Recall basic capacitance relations.
For a parallel plate capacitor, \[ C = \frac{\varepsilon_0 k A}{d} \] Capacitance increases with higher dielectric constant and decreases with effective separation.
Step 2: Analyze configuration (A).
In case (A), dielectrics \(k_1\) and \(k_2\) are arranged partially in series and partially in parallel. Due to larger contribution of lower dielectric \(k_2\) in series combination, the overall capacitance becomes minimum among the three. Hence, \(C_A\) is the smallest.
Step 3: Analyze configuration (B).
In case (B), both dielectrics are symmetrically arranged such that the effective dielectric contribution is maximum. The higher dielectric \(k_1\) dominates more effectively, giving the largest equivalent capacitance. Hence, \(C_B\) is maximum.
Step 4: Analyze configuration (C).
In case (C), the dielectrics are arranged side by side leading to a parallel combination of capacitors. The equivalent capacitance lies between cases (A) and (B). Thus, \[ C_B>C_C>C_A \]
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