The plates of a parallel plate capacitor are separated by d. Two… | NEET 2025

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Dielectrics

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 $K_1 = 1.25K_2$ , the value of $K_1$ is:

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Answer: (c)

Air gap left $=d-\frac{3d}{8}-\frac{d}{2}=\frac{d}{8}$ . All in series: $\frac{d}{8}+\frac{3d}{8K_1}+\frac{d}{2K_2}=\frac{d}{2}$
$K_2=\frac{K_1}{1.25} \Rightarrow \frac{1}{2K_2}=\frac{0.625}{K_1}$
$\frac{0.375+0.625}{K_1}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8} \Rightarrow K_1=\frac{8}{3}$
So the answer is (c) 2.66.

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