JEE Main 2025 Apr 3 Shift 1, Physics Q8: Capacitance For Geometrical Figures With Dielectric

JEE Main2025Apr 3, Shift 1Physics

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 :

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

Solution

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

First configuration (series, horizontal slabs): Each half-gap has thickness $d /2$ with full area $A$ .

$C_{upper} = \frac{ε _{2} ε _{0} A}{d /2} = 2 ε_{2} C_{0}$ , $C_{lower} = 2 ε_{1} C_{0} .$

In series: $C_{1} = \frac{( 2 ε _{1} C _{0} ) ( 2 ε _{2} C _{0} )}{2 ε _{1} C _{0} + 2 ε _{2} C _{0}} = \frac{2 ε _{1} ε _{2} C _{0}}{ε _{1} + ε _{2}} .$

Second configuration (parallel, vertical slabs): Each half-area $A /2$ spans the full gap $d$ .

$C_{left} = \frac{ε _{2} ε _{0} ( A /2 )}{d} = \frac{ε _{2} C _{0}}{2}$ , $C_{right} = \frac{ε _{1} C _{0}}{2} .$

In parallel: $C_{2} = \frac{( ε _{1} + ε _{2} ) C _{0}}{2} .$

Ratio: $\frac{C _{1}}{C _{2}} = \frac{2 ε _{1} ε _{2} C _{0}}{ε _{1} + ε _{2}} \cdot \frac{2}{( ε _{1} + ε _{2} ) C _{0}} = \frac{4 ε _{1} ε _{2}}{( ε _{1} + ε _{2} ) ^{2}} .$

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JEE Main 2025 Apr 3 Shift 1, Physics Q8: Capacitance For Geometrical Figures With Dielectric

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