NEET 2025 — Physics solution

NEET2025Physics

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:

A
2.66
B
2.33
C
1.60
D
1.33

Solution

Original capacitance: $C_{1} = ϵ_{0} A / d$ .

With two dielectric slabs (thicknesses $3 d /8$ and $d /2$ ) and an air gap of thickness $d - 3 d /8 - d /2 = d /8$ , in series:

$\frac{1}{C _{2}} = \frac{3 d /8}{K _{1} ϵ _{0} A} + \frac{d /2}{K _{2} ϵ _{0} A} + \frac{d /8}{ϵ _{0} A}$ .

$C_{2} = \frac{ϵ _{0} A / d}{3/ ( 8 K _{1} ) + 1/ ( 2 K _{2} ) + 1/8}$ . Setting $C_{2} = 2 C_{1}$ and using $K_{2} = K_{1} /1.25 = 4 K_{1} /5$ :

$\frac{3}{8 K _{1}} + \frac{5}{8 K _{1}} + \frac{1}{8} = \frac{1}{2} \Rightarrow \frac{1}{K _{1}} + \frac{1}{8} = \frac{1}{2} \Rightarrow \frac{1}{K _{1}} = \frac{3}{8} \Rightarrow K_{1} = \frac{8}{3} \approx 2.66$ .

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