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JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 9)

A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is $d$ and area of each plate is $A$. If capacitance in first configuration and second configuration are $\mathrm{C}_1$ and $\mathrm{C}_2$ respectively, then $\frac{C_1}{C_2}$ is:

First Configuration

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Capacitor Question 4 English 1

Second Configuration

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Capacitor Question 4 English 2

$\frac{\varepsilon_0\left(\varepsilon_1+\varepsilon_2\right)}{2}$
$\frac{\varepsilon_1 \varepsilon_2^2}{\left(\varepsilon_1+\varepsilon_2\right)^2}$
$\frac{4 \varepsilon_1 \varepsilon_2}{\left(\varepsilon_1+\varepsilon_2\right)^2}$
$\frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1+\varepsilon_2}$

Explanation

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Capacitor Question 4 English Explanation 1

Area of plate is A. then

$$\begin{aligned} & \mathrm{C}=\frac{\varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \\ & \mathrm{C}^{\prime}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \end{aligned}$$

$$\begin{aligned} &\text { Let } \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\\ &\mathrm{C}=2 \varepsilon_2 \mathrm{C}_0\\ &\mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0 \end{aligned}$$

C & C' are in series

$$\begin{aligned} & \mathrm{C}_1=\frac{\mathrm{CC}^{\prime}}{\mathrm{C}+\mathrm{C}^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 \mathrm{C}_0^2}{2 \mathrm{C}_0\left(\varepsilon_2+\varepsilon_1\right)} \\ & =\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \end{aligned}$$

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Capacitor Question 4 English Explanation 2

Here $C=\frac{\varepsilon_1 \varepsilon_0 A}{2 d}=\frac{\varepsilon_1 C_0}{2}$

$$C^{\prime}=\frac{\varepsilon_2 C_0}{2}$$

$\mathrm{C} \& \mathrm{C}^{\prime}$ are inparallel

$$\mathrm{C}_2=\mathrm{C}^{\prime}+\mathrm{C}=\left(\varepsilon_1+\varepsilon_2\right) \frac{\mathrm{C}_0}{2}$$

Thus $\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \times \frac{2}{\left(\varepsilon_1+\varepsilon_2\right) \mathrm{C}_0}$

$$=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1\right)^2}$$

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