JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

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? $(k_{1} > k_{2})$

A
$C_{B} > C_{C} > C_{A}$
B
$C_{C} > C_{B} > C_{A}$
C
$C_{C} > C_{A} > C_{B}$
D
$C_{A} > C_{C} > C_{B}$

Solution

Let $C = \frac{ε _{0} A}{d}$ be the reference capacitance.

Configuration A (top half all $k_{1}$ ; bottom half $k_{1}$ left, $k_{2}$ right): the top half is a single capacitor with $k_{1}$ ; the bottom half is two capacitors $k_{1}$ and $k_{2}$ side-by-side (parallel). These two halves stack in series.

$C_{A} = K_{1} C \cdot \frac{K _{1} + 2 K _{2}}{2 ( K _{1} + K _{2} )}$ (after combining).

Configuration B (top half all $k_{2}$ ; bottom half $k_{1}$ left, $k_{2}$ right): symmetric to A but with $k_{2}$ replacing $k_{1}$ on top.

$C_{B} = K_{2} C \cdot \frac{K _{1} + 2 K _{2}}{2 ( K _{1} + K _{2} )}$ .

Configuration C (diagonal $k_{1}, k_{2}$ on top; $k_{2}, k_{1}$ on bottom): each vertical half is a series combination of $k_{1}$ and $k_{2}$ , giving $\frac{2 K _{1} K _{2}}{K _{1} + K _{2}} \cdot \frac{C}{2}$ each; two such halves in parallel.

$C_{C} = \frac{2 K _{1} K _{2} C}{K _{1} + K _{2}}$ .

Since $k_{1} > k_{2}$ , multiplying by $K_{1}$ vs $K_{2}$ in the prefactor of $C_{A}, C_{B}$ gives $C_{A} > C_{B}$ . Comparing arithmetic and harmonic means: $C_{A} > C_{C} > C_{B}$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →