A parallel plate capacitor has capacitance C , when there is vacuum within the parallel plates. A sheet having thickness $\left(\frac{1}{3}\right)^{\text {rd }}$ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

(A) $\frac{3 \mathrm{KC}}{2 \mathrm{~K}+1}$

(B) $\frac{\text { CK }}{2+\text { K }}$

(C) $\frac{3 \mathrm{CK}^{2}}{(2 \mathrm{~K}+1)^{2}}$

(D) $\frac{4 \mathrm{KC}}{3 \mathrm{~K}-1}$

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Capacitance

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

(\frac{1}{3})^{rd}

of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

Answer: (A)

The slab of thickness

d /3

and the remaining vacuum gap

2 d /3

behave as two capacitors in series. The vacuum part has

C_{1} = \frac{ϵ _{0} A}{2 d /3} = \frac{3 C}{2}

and the dielectric part has

C_{2} = \frac{K ϵ _{0} A}{d /3} = 3 K C

. Combining in series,

C^{'} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{\frac{3 C}{2} \cdot 3 K C}{\frac{3 C}{2} + 3 K C} = \frac{3 K C}{2 K + 1}

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option