JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Capacitance

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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)

(A) $C_{B} > C_{C} > C_{A}$
(B) $C_{C} > C_{B} > C_{A}$
(C) $C_{C} > C_{A} > C_{B}$
(D) $C_{A} > C_{C} > C_{B}$ ✓ Correct

Answer: (D)

Solution

Writing

C = \frac{ε _{0} A}{d}

, capacitor

A

has an upper slab of thickness

d /2

filled with

k_{1}

over the full plate, giving

2 k_{1} C

, in series with a lower slab (thickness

d /2

) whose two halves

k_{1}

and

k_{2}

sit side by side, giving

(k_{1} + k_{2}) C

; hence

C_{A} = \frac{2 k _{1} ( k _{1} + k _{2} )}{3 k _{1} + k _{2}} C

. Capacitor

B

is the same with

k_{2}

as the full upper slab, so

C_{B} = \frac{2 k _{2} ( k _{1} + k _{2} )}{k _{1} + 3 k _{2}} C

, and capacitor

C

has the two dielectrics stacked across the whole gap in series,

C_{C} = \frac{2 k _{1} k _{2}}{k _{1} + k _{2}} C

. With

k_{1} > k_{2}

, forming the differences shows

C_{A} - C_{C} > 0

and

C_{C} - C_{B} > 0

. Therefore

C_{A} > C_{C} > C_{B}

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Capacitance

Solution

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