The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K 1 and K 2 with thickness 3/8 d and d/2 , respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. (If K 1 =1.25 K 2 , the value of K 1 is :

Physics · Capacitors and R-C Circuits

NEET (UG) 2025 — Question 19

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant $K_{1}$ and $K_{2}$ with thickness $\frac{3}{8} d$ and $\frac{d}{2}$ , respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. (If $K_{1}=1.25 K_{2}$ , the value of $K_{1}$ is :

Option A:
2.66
Correct
Option B:
2.33
Option C:
1.6
Option D:
1.33

Answer: A

Step-by-step solution

$C_1 = \epsilon_0 A/d$

$C_2 = \frac{\epsilon_0 A}{\frac{3d}{8K_1} + \frac{d}{2K_2} + (d - \frac{3d}{8} - \frac{d}{2})}$

$C_2 = \frac{\epsilon_0 A/d}{\frac{3}{8K_1} + \frac{1}{2K_2} + \frac{1}{8}}$

$= \frac{\epsilon_0 A/d}{\frac{3}{8K_1} + \frac{1}{2}\left(\frac{4}{5K_1}\right) + \frac{1}{8}}$

$C_2 = 2C_1$

$\frac{\epsilon_0 A/d}{\frac{3}{8K_1} + \frac{5}{8K_1} + \frac{1}{8}} = 2\epsilon_0 A/d$

$\implies \frac{1}{\frac{1}{K_1} + \frac{1}{8}} = 2 \implies \frac{1}{K_1} + \frac{1}{8} = \frac{1}{2} \implies K_1 = \frac{8}{3} = 2.66$

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2025
Paper: 2025 paper
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Effect of Dielectrics