JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates.

A sheet having thickness $(\frac{1}{3})^{r d}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

A
$\frac{3 K C}{2 K + 1}$
B
$\frac{C K}{2 + K}$
C
$\frac{3 C K ^{2}}{( 2 K + 1 ) ^{2}}$
D
$\frac{4 K C}{3 K - 1}$

Solution

Original: $C = \frac{ε _{0} A}{d}$ . With the dielectric sheet (thickness $d /3$ ) inserted, the system is equivalent to two capacitors in series:

$C_{1}$ (vacuum, gap $2 d /3$ ): $C_{1} = \frac{ε _{0} A}{2 d /3} = \frac{3}{2} C$ .

$C_{2}$ (dielectric, gap $d /3$ ): $C_{2} = \frac{K ε _{0} A}{d /3} = 3 K C$ .

Series combination: $C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{( 3/2 ) C \cdot 3 K C}{( 3/2 ) C + 3 K C} = \frac{( 9/2 ) K C}{( 3/2 ) ( 1 + 2 K )} = \frac{3 K C}{2 K + 1}$ .

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