Question 193: Electrostatics & 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
![](['https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m9gibs63/ce4fae89-aa3b-45ff-a7f5-c470eaa3d061/906272b0-18df-11f0-a103-9bd1ffe79f48/file-jaoe38c1m9gibs64.png?format=png', 'https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m9giblba/25fc882b-41c0-436f-8978-fc0913f4b9b4/8b16b460-18df-11f0-a103-9bd1ffe79f48/file-jaoe38c1m9giblbb.png?format=png'])

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 $\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1ight)} 	imes \frac{2}{\left(\varepsilon_1+\varepsilon_2ight) \mathrm{C}_0}$
$=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1ight)^2}$
![](['https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1ma0jbf4k/355fbf12-9837-4d84-a3d7-59c6bb3c4381/1c6d5140-23e3-11f0-a048-651b25b521d4/file-jaoe38c1ma0jbf4l.png?format=png', 'https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1ma0jdwsn/2cf3fa9b-f1c9-4228-8c89-b39f79340007/61b1b570-23e3-11f0-a048-651b25b521d4/file-jaoe38c1ma0jdwso.png?format=png'])

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

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

Electrostatics & Capacitance

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