Physics Q38 – JEE-MAIN 2025

Home Physics Electrostatics & CapacitanceQuestion #131

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

ϵ_{1}

and

ϵ_{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}}

Clue 1 of 3: Identify the configuration

For the first configuration, the dielectrics are in series, and for the second, they are in parallel.

🥷

Ninja Strategy•Symmetry and Dimensional Analysis

First, eliminate options that are dimensionally incorrect for a ratio. Then, test the remaining options by setting $ϵ_{1} = ϵ_{2} = ϵ$ , which should yield a ratio of 1.

🧠 Full Solution Path

Step 1: Calculate Capacitance for First Configuration (

C_{1}

)✦ Active

In the first configuration, the two dielectrics are in series. Each dielectric fills half the distance, $d / 2$ . The capacitance for each section is $C_{1}^{'} = \frac{ϵ_{1} ϵ_{0} A}{d / 2} = \frac{2 ϵ_{1} ϵ_{0} A}{d}$ and $C_{2}^{'} = \frac{ϵ_{2} ϵ_{0} A}{d / 2} = \frac{2 ϵ_{2} ϵ_{0} A}{d}$ . The equivalent capacitance $C_{1}$ for series combination is:

\frac{1}{C_{1}} = \frac{1}{C_{1}^{'}} + \frac{1}{C_{2}^{'}} = \frac{d}{2 ϵ_{1} ϵ_{0} A} + \frac{d}{2 ϵ_{2} ϵ_{0} A} = \frac{d}{2 ϵ_{0} A} (\frac{1}{ϵ_{1}} + \frac{1}{ϵ_{2}}) = \frac{d}{2 ϵ_{0} A} \frac{ϵ_{1} + ϵ_{2}}{ϵ_{1} ϵ_{2}}

Therefore,

C_{1} = \frac{2 ϵ_{1} ϵ_{2} ϵ_{0} A}{d (ϵ_{1} + ϵ_{2})}

Step 2: Calculate Capacitance for Second Configuration (

C_{2}

)○ Expand

In the second configuration, the two dielectrics are in parallel. Each dielectric fills half the area, $A / 2$ . The capacitance for each section is $C_{1}^{″} = \frac{ϵ_{1} ϵ_{0} (A / 2)}{d} = \frac{ϵ_{1} ϵ_{0} A}{2 d}$ and $C_{2}^{″} = \frac{ϵ_{2} ϵ_{0} (A / 2)}{d} = \frac{ϵ_{2} ϵ_{0} A}{2 d}$ . The equivalent capacitance $C_{2}$ for parallel combination is:

C_{2} = C_{1}^{″} + C_{2}^{″} = \frac{ϵ_{1} ϵ_{0} A}{2 d} + \frac{ϵ_{2} ϵ_{0} A}{2 d} = \frac{(ϵ_{1} + ϵ_{2}) ϵ_{0} A}{2 d}

Step 3: Calculate the Ratio

\frac{C_{1}}{C_{2}}

○ Expand

Now, we find the ratio of $C_{1}$ to $C_{2}$ :

\frac{C_{1}}{C_{2}} = \frac{\frac{2 ϵ_{1} ϵ_{2} ϵ_{0} A}{d (ϵ_{1} + ϵ_{2})}}{\frac{(ϵ_{1} + ϵ_{2}) ϵ_{0} A}{2 d}} = \frac{2 ϵ_{1} ϵ_{2} ϵ_{0} A}{d (ϵ_{1} + ϵ_{2})} \times \frac{2 d}{(ϵ_{1} + ϵ_{2}) ϵ_{0} A}

Simplifying the expression, we get:

\frac{C_{1}}{C_{2}} = \frac{4 ϵ_{1} ϵ_{2}}{(ϵ_{1} + ϵ_{2})^{2}}

💡 Teacher's Secret Hint

Ensure all common terms like $ϵ_{0} A / d$ cancel out correctly.

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Physics Question 38 – JEE-MAIN 2025

Clue 1 of 3: Identify the configuration

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