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Capacitor question

2025 · Q177
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Capacitor question

2025 · Q177

NEETPhysicsCapacitorMCQ+4 / −1
The plates of a parallel plate capacitor are separated by ddd. Two slabs of different dielectric constant K1K_1K1​ and K2K_2K2​ with thickness 38d\frac{3}{8} d83​d and d2\frac{d}{2}2d​, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1=1.25K2K_1=1.25 K_2K1​=1.25K2​, the value of K1K_1K1​ is:
  1. A
    1.60
  2. B
    1.33
  3. C
    2.66
  4. D
    2.33
View written solutionFree

Correct answer: C

NEET 2025 Physics - Capacitor Question 1 English Explanation

Using $C_{e q}=\frac{\varepsilon_0 A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}}$

here $C_0=\frac{\varepsilon_0 A}{d}, t_1=\frac{3 d}{8}, t_2=\frac{d}{2}, t_3=\frac{d}{8}$

$$K_1=K_1, \quad K_2=\frac{K_1}{1.25} \text { and } K_3=1$$

Given $C_{\text {eq }}=2 C_0$

$$\begin{aligned} \& \Rightarrow 2 C_0=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d \times 1.25}{2 k_1}+\frac{d}{8}} \\ \& \Rightarrow \frac{2 \varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d}{2 k_1} \times \frac{5}{4}+\frac{d}{8}} \\ \& \Rightarrow 2=\frac{1}{\frac{3}{8 k_1}+\frac{5}{8 k_1}+\frac{1}{8}} \Rightarrow k_1=\frac{8}{3}=2.66 \end{aligned}$$

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