JEE Main 2025 — Physics Electromagnetism

Area of plate is $A$A. then

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

Let $\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$C0​= dε0​ A​

$\begin{aligned}
& \mathrm{C}=2 \varepsilon_2 \mathrm{C}_0 \
& \mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0
\end{aligned}$

$\mathrm{C} \& \mathrm{C}^{\prime}$C&C′ are in series

$$\mathrm{C}_1=\frac{\mathrm{CC}^{\prime}}{\mathrm{C}+\mathrm{C}^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 \mathrm{C}_0^2}{2 \mathrm{C}_0\left(\varepsilon_2+\varepsilon_1\right)}$$C1​=C+C′CC′​=2C0​(ε2​+ε1​)4ε2​ε1​C02​​

$=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)}$=(ε2​+ε1​)2ε2​ε1​C0​​

Here $\mathrm{C}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=\frac{\varepsilon_1 \mathrm{C}_0}{2}$C=2 dε1​ε0​ A​=2ε1​C0​​

$$\mathrm{C}^{\prime}=\frac{\varepsilon_2 \mathrm{C}_0}{2}$$C′=2ε2​C0​​

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

$$\mathrm{C}_2=\mathrm{C}^{\prime}+\mathrm{C}=\left(\varepsilon_1+\varepsilon_2\right) \frac{\mathrm{C}_0}{2}$$C2​=C′+C=(ε1​+ε2​)2C0​​

Thus $\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \times \frac{2}{\left(\varepsilon_1+\varepsilon_2\right) \mathrm{C}_0}$C2​C1​​=(ε2​+ε1​)2ε2​ε1​C0​​×(ε1​+ε2​)C0​2​

$$=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1\right)^2}$$=(ε2​+ε1​)24ε2​ε1​​