Q53 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
A parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. A sheet having thickness $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of the separation between the plates and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is :
- A$\frac{4 K C}{3 K-1}$
- B$\frac{3 C K^{2}}{(2 K+1)^{2}}$
- Correct answerC$\frac{3 K C}{2 K+1}$
- D$\frac{C K}{2+K}$
Solution
Original capacitance: $C = \frac{\varepsilon_0 A}{d}$
After inserting dielectric of thickness $\frac{d}{3}$, system becomes two capacitors in series.
Vacuum region: $C_1 = \frac{\varepsilon_0 A}{2d/3} = \frac{3C}{2}$
Dielectric region: $C_2 = \frac{K\varepsilon_0 A}{d/3} = 3KC$
Equivalent capacitance: $\frac{1}{C_{new}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{2}{3C} + \frac{1}{3KC} = \frac{2K+1}{3KC}$
$C_{new} = \frac{3KC}{2K+1}$
After inserting dielectric of thickness $\frac{d}{3}$, system becomes two capacitors in series.
Vacuum region: $C_1 = \frac{\varepsilon_0 A}{2d/3} = \frac{3C}{2}$
Dielectric region: $C_2 = \frac{K\varepsilon_0 A}{d/3} = 3KC$
Equivalent capacitance: $\frac{1}{C_{new}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{2}{3C} + \frac{1}{3KC} = \frac{2K+1}{3KC}$
$C_{new} = \frac{3KC}{2K+1}$
Hence, the correct answer is option: 'C'.