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 C_1 and C_2 respectively, then \fracC_1C_2 is: First Configuration Second Configuration

\frac\varepsilon_0(\varepsilon_1+\varepsilon_2)2

Capacitor PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q59 with Solution

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Correct answer: $\DFRAC{C_1}{C_2} = \DFRAC{(\VAREPSILON_1+\VAREPSILON_2)^2}{4\VAREPSILON_1\VAREPSILON_2}$, IF THE QUESTION INTENDED $\DFRAC{C_2}{C_1}$, THEN OPTION C WOULD BE CORRECT.

Interpret the two configurations

For a parallel plate capacitor, capacitance depends on how the dielectrics are arranged.

First configuration: the dielectric slab is divided into two equal parts along the area. So each dielectric occupies area $A/2$ and thickness $d$ . This is equivalent to two capacitors in parallel.
Second configuration: the dielectric slab is divided into two equal parts along the thickness. So each dielectric occupies thickness $d/2$ and area $A$ . This is equivalent to two capacitors in series.

Capacitance in first configuration: $C_1$

Each part has area $A/2$ and plate separation $d$ .

So,

C_{1a} = \frac{\varepsilon_0 \varepsilon_1 (A/2)}{d} = \frac{\varepsilon_0 \varepsilon_1 A}{2d}

C_{1b} = \frac{\varepsilon_0 \varepsilon_2 (A/2)}{d} = \frac{\varepsilon_0 \varepsilon_2 A}{2d}

Since they are in parallel,

C_1 = C_{1a} + C_{1b}

C_1 = \frac{\varepsilon_0 A}{2d}(\varepsilon_1 + \varepsilon_2)

Capacitance in second configuration: $C_2$

Now the two dielectric layers are stacked along the thickness, each of thickness $d/2$ .

Capacitance of first layer:

C_{2a} = \frac{\varepsilon_0 \varepsilon_1 A}{d/2} = \frac{2\varepsilon_0 \varepsilon_1 A}{d}

Capacitance of second layer:

C_{2b} = \frac{\varepsilon_0 \varepsilon_2 A}{d/2} = \frac{2\varepsilon_0 \varepsilon_2 A}{d}

These are in series, so

\frac{1}{C_2} = \frac{1}{C_{2a}} + \frac{1}{C_{2b}}

\frac{1}{C_2} = \frac{d}{2\varepsilon_0 \varepsilon_1 A} + \frac{d}{2\varepsilon_0 \varepsilon_2 A}

\frac{1}{C_2} = \frac{d}{2\varepsilon_0 A}\left(\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2}\right)

\frac{1}{C_2} = \frac{d}{2\varepsilon_0 A}\cdot \frac{\varepsilon_1+\varepsilon_2}{\varepsilon_1\varepsilon_2}

Hence,

C_2 = \frac{2\varepsilon_0 A}{d}\cdot \frac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2}

Find the ratio $\dfrac{C_1}{C_2}$

\frac{C_1}{C_2} = \frac{\dfrac{\varepsilon_0 A}{2d}(\varepsilon_1+\varepsilon_2)}{\dfrac{2\varepsilon_0 A}{d}\cdot \dfrac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2}}

Cancel $\varepsilon_0 A/d$ :

\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)/2}{2\varepsilon_1\varepsilon_2/(\varepsilon_1+\varepsilon_2)}

\frac{C_1}{C_2} = \frac{\varepsilon_1+\varepsilon_2}{2} \cdot \frac{\varepsilon_1+\varepsilon_2}{2\varepsilon_1\varepsilon_2}

\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)^2}{4\varepsilon_1\varepsilon_2}

Compare with options

The derived ratio is

\boxed{\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)^2}{4\varepsilon_1\varepsilon_2}}

But option C is

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

which is actually the reciprocal:

\frac{C_2}{C_1}

So the stored answer appears reversed.

Final conclusion

\boxed{\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)^2}{4\varepsilon_1\varepsilon_2}}

None of the listed options matches this expression; option C corresponds to $\dfrac{C_2}{C_1}$ , not $\dfrac{C_1}{C_2}$ .

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View written solutionFree

Correct answer: $\DFRAC{C_1}{C_2} = \DFRAC{(\VAREPSILON_1+\VAREPSILON_2)^2}{4\VAREPSILON_1\VAREPSILON_2}$, IF THE QUESTION INTENDED $\DFRAC{C_2}{C_1}$, THEN OPTION C WOULD BE CORRECT.

Interpret the two configurations

For a parallel plate capacitor, capacitance depends on how the dielectrics are arranged.

First configuration: the dielectric slab is divided into two equal parts along the area. So each dielectric occupies area $A/2$ and thickness $d$ . This is equivalent to two capacitors in parallel.
Second configuration: the dielectric slab is divided into two equal parts along the thickness. So each dielectric occupies thickness $d/2$ and area $A$ . This is equivalent to two capacitors in series.

Capacitance in first configuration: $C_1$

Each part has area $A/2$ and plate separation $d$ .

So,

C_{1a} = \frac{\varepsilon_0 \varepsilon_1 (A/2)}{d} = \frac{\varepsilon_0 \varepsilon_1 A}{2d}

C_{1b} = \frac{\varepsilon_0 \varepsilon_2 (A/2)}{d} = \frac{\varepsilon_0 \varepsilon_2 A}{2d}

Since they are in parallel,

C_1 = C_{1a} + C_{1b}

C_1 = \frac{\varepsilon_0 A}{2d}(\varepsilon_1 + \varepsilon_2)

Capacitance in second configuration: $C_2$

Now the two dielectric layers are stacked along the thickness, each of thickness $d/2$ .

Capacitance of first layer:

C_{2a} = \frac{\varepsilon_0 \varepsilon_1 A}{d/2} = \frac{2\varepsilon_0 \varepsilon_1 A}{d}

Capacitance of second layer:

C_{2b} = \frac{\varepsilon_0 \varepsilon_2 A}{d/2} = \frac{2\varepsilon_0 \varepsilon_2 A}{d}

These are in series, so

\frac{1}{C_2} = \frac{1}{C_{2a}} + \frac{1}{C_{2b}}

\frac{1}{C_2} = \frac{d}{2\varepsilon_0 \varepsilon_1 A} + \frac{d}{2\varepsilon_0 \varepsilon_2 A}

\frac{1}{C_2} = \frac{d}{2\varepsilon_0 A}\left(\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2}\right)

\frac{1}{C_2} = \frac{d}{2\varepsilon_0 A}\cdot \frac{\varepsilon_1+\varepsilon_2}{\varepsilon_1\varepsilon_2}

Hence,

C_2 = \frac{2\varepsilon_0 A}{d}\cdot \frac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2}

Find the ratio $\dfrac{C_1}{C_2}$

\frac{C_1}{C_2} = \frac{\dfrac{\varepsilon_0 A}{2d}(\varepsilon_1+\varepsilon_2)}{\dfrac{2\varepsilon_0 A}{d}\cdot \dfrac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2}}

Cancel $\varepsilon_0 A/d$ :

\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)/2}{2\varepsilon_1\varepsilon_2/(\varepsilon_1+\varepsilon_2)}

\frac{C_1}{C_2} = \frac{\varepsilon_1+\varepsilon_2}{2} \cdot \frac{\varepsilon_1+\varepsilon_2}{2\varepsilon_1\varepsilon_2}

\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)^2}{4\varepsilon_1\varepsilon_2}

Compare with options

The derived ratio is

\boxed{\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)^2}{4\varepsilon_1\varepsilon_2}}

But option C is

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

which is actually the reciprocal:

\frac{C_2}{C_1}

So the stored answer appears reversed.

Final conclusion

\boxed{\frac{C_1}{C_2} = \frac{(\varepsilon_1+\varepsilon_2)^2}{4\varepsilon_1\varepsilon_2}}

None of the listed options matches this expression; option C corresponds to $\dfrac{C_2}{C_1}$ , not $\dfrac{C_1}{C_2}$ .

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