JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Capacitance

A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant

ε_{1}

and

ε_{2}

, as shown in figures. The distance between the plates is d and area of each plate is A . If capacitance in first configuration and second configuration are

C_{1}

and

C_{2}

respectively, then

\frac{C _{1}}{C _{2}}

is :

Structure Configuration 1 — Configuration 1

Structure Configuration 2 — Configuration 2

(A) $\frac{ε _{1} ε _{2}^{2}}{( ε _{1} + ε _{2} ) ^{2}}$
(B) $\frac{4 ε _{1} ε _{2}}{( ε _{1} + ε _{2} ) ^{2}}$ ✓ Correct
(C) $\frac{ε _{1} ε _{2}}{ε _{1} + ε _{2}}$
(D) $\frac{ε _{0} ( ε _{1} + ε _{2} )}{2}$

Answer: (B)

Solution

Let

C_{0} = \frac{ε _{0} A}{d}

. In the first configuration the two dielectric slabs are stacked, each filling a gap

d /2

over the full area, forming two capacitors in series:

C_{a} = \frac{ε _{1} ε _{0} A}{d /2} = 2 ε_{1} C_{0}, C_{b} = 2 ε_{2} C_{0},

C_{1} = \frac{C _{a} C _{b}}{C _{a} + C _{b}} = \frac{2 ε _{1} ε _{2}}{ε _{1} + ε _{2}} C_{0} .

In the second configuration each dielectric fills half the area over the full gap

d

, giving two capacitors in parallel:

C_{2} = \frac{ε _{1} C _{0}}{2} + \frac{ε _{2} C _{0}}{2} = \frac{ε _{1} + ε _{2}}{2} C_{0} .

Therefore

\frac{C _{1}}{C _{2}} = \frac{2 ε _{1} ε _{2} / ( ε _{1} + ε _{2} )}{( ε _{1} + ε _{2} ) /2} = \frac{4 ε _{1} ε _{2}}{( ε _{1} + ε _{2} ) ^{2}} .

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Capacitance

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