Question 124: Electrostats & Capacitance

Question

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

Second Configuration

Accepted Answer

Area of plate is A. then
$$\begin{aligned}
& \mathrm{C}=\frac{\varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \
& \mathrm{C}^{\prime}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}
\end{aligned}$$
$$\begin{aligned}
&	ext { Let } \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\
&\mathrm{C}=2 \varepsilon_2 \mathrm{C}_0\
&\mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0
\end{aligned}$$
C & C' are in series
$$\begin{aligned}
& \mathrm{C}_1=\frac{\mathrm{CC}^{\prime}}{\mathrm{C}+\mathrm{C}^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 \mathrm{C}_0^2}{2 \mathrm{C}_0\left(\varepsilon_2+\varepsilon_1ight)} \
& =\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1ight)}
\end{aligned}$$
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
$$\mathrm{C}_2=\mathrm{C}^{\prime}+\mathrm{C}=\left(\varepsilon_1+\varepsilon_2ight) \frac{\mathrm{C}_0}{2}$$
Thus $\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1ight)} 	imes \frac{2}{\left(\varepsilon_1+\varepsilon_2ight) \mathrm{C}_0}$
$$=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1ight)^2}$$

Answer

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

Answer

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

Answer

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

Answer

$\frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1+\varepsilon_2}$

Electrostats & Capacitance

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