A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $\varepsilon_1$ 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 $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is :

Question

Aicharya · Accepted Answer

Area of plate is $A$. then
$\begin{aligned}
& \mathrm{C}=\frac{\varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \\ & \mathrm{C}^{\prime}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}
\end{aligned}$
Let $\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$
$\begin{aligned}
& \mathrm{C}=2 \varepsilon_2 \mathrm{C}_0 \\ & \mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0
\end{aligned}$
$\mathrm{C} \& \mathrm{C}^{\prime}$ are in series
$\mathrm{C}_1=\frac{\mathrm{CC}^{\prime}}{\mathrm{C}+\mathrm{C}^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 \mathrm{C}_0^2}{2 \mathrm{C}_0\left(\varepsilon_2+\varepsilon_1\right)}$
$=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)}$

Here $\mathrm{C}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=\frac{\varepsilon_1 \mathrm{C}_0}{2}$
$\mathrm{C}^{\prime}=\frac{\varepsilon_2 \mathrm{C}_0}{2}$
$\mathrm{C} \& \mathrm{C}^{\prime}$ are inparallel
$\mathrm{C}_2=\mathrm{C}^{\prime}+\mathrm{C}=\left(\varepsilon_1+\varepsilon_2\right) \frac{\mathrm{C}_0}{2}$
Thus $\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \times \frac{2}{\left(\varepsilon_1+\varepsilon_2\right) \mathrm{C}_0}$
$=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1\right)^2}$

A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant…

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