A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant _1 and _2 , as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first

Correct answer: B. (4 _1 _2)/( ( _1+ _2)^2) Area of plate is A . then aligned & C= _2 _0 ~A ~d / 2 = 2 _2 _0 ~A ~d & C^ = _1 _0 ~A ~d / 2 = 2 _1 _0 ~A ~d aligned Let C_0= _0 ~A ~d aligned & C=2 _2 C_0 & C^ =2 _1 C_0 aligned C \& C^ are in series C_1= CC^ C+C^ = 4 _2 _1 C_0^2 2 C_0( _2+ _1) = 2 _2 _1 C_0 ( _2+ _1) Here C= _1 _0 ~A 2 ~d = _1 C_0 2 C^ = _2 C_0 2 C \& C^ are inparallel C_2=C^ +C=( _1+ _2) C_0 2 Thus C_1 C_2 = 2 _2 _1 C_0 ( _2+ _1) × 2 ( _1+ _2) C_0 =(4 _2 _1)/(( _2+ _1)^2)

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Physics
Chapter: Capacitance

JEE Main 2025 · 3 April · Morning Shift · Physics · Capacitance

A parallel plate capacitor is filled equally (half) with two...

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 :

A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant _1 and _2 , as shown in

Options

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

Correct answer

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

Step-by-step solution

Area of plate is

A

. then

\begin{aligned}<br/>& \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}}<br/>\end{aligned}

Let

\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}

\begin{aligned}<br/>& \mathrm{C}=2 \varepsilon_2 \mathrm{C}_0 \\ & \mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0<br/>\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}

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