oads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/18f70d92eaab.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="NEET 2025 Physics - Capacitor Question 4 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 $C_{\text {eq }}=2 C_0$

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

19:T4ac,

NEET 2025 Physics - Capacitor Question 4 Hindi Explanation

$C_{e q}=\frac{\varepsilon_0 A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}}$ का उपयोग करते हुए

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

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

दिया गया है $C_{\text {eq }}=2 C_0$

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