A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

\varepsilon _0A \over d( 1 \over 2 + K_1K_2 \over K_1 + K_2 )

\varepsilon _0A \over d( 1 \over 2 + K_1K_2 \over 2(K_1 + K_2) )

\varepsilon _0A \over d( 1 \over 2 + K_1 + K_2 \over K_1K_2 )

\varepsilon _0A \over d( 1 \over 2 + 2(K_1 + K_2) \over K_1K_2 )

Capacitor PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q55 with Solution

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Correct answer: A

Interpret the arrangement

The capacitor has plate area $A$ and separation $d$ .

Two dielectric slabs are inserted, each having:

area $A/2$
thickness $d/2$
dielectric constants $K_1$ and $K_2$

This means the capacitor is effectively divided into two parallel parts, each of area $A/2$ .

In one half-area, the space is filled by two layers along the thickness: one layer of dielectric and one air layer.
In the other half-area, similarly the second dielectric occupies half the thickness and the remaining half is air.

So we get two capacitors in parallel:

Left half: area $A/2$ , with dielectric $K_1$ of thickness $d/2$ in series with air of thickness $d/2$
Right half: area $A/2$ , with dielectric $K_2$ of thickness $d/2$ in series with air of thickness $d/2$

Capacitance of first half

For the first half-area $A/2$ :

Capacitance of dielectric part:

C_{11} = \frac{\varepsilon_0 K_1 (A/2)}{d/2} = \frac{\varepsilon_0 K_1 A}{d}

Capacitance of air part:

C_{12} = \frac{\varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d}

These are in series, so

\frac{1}{C_1} = \frac{1}{C_{11}} + \frac{1}{C_{12}} = \frac{d}{\varepsilon_0 K_1 A} + \frac{d}{\varepsilon_0 A}

\frac{1}{C_1} = \frac{d}{\varepsilon_0 A}\left(\frac{1}{K_1}+1\right) = \frac{d}{\varepsilon_0 A}\cdot \frac{K_1+1}{K_1}

Hence,

C_1 = \frac{\varepsilon_0 A}{d}\cdot \frac{K_1}{K_1+1}

Capacitance of second half

Similarly, for dielectric constant $K_2$ :

C_{21} = \frac{\varepsilon_0 K_2 (A/2)}{d/2} = \frac{\varepsilon_0 K_2 A}{d}

C_{22} = \frac{\varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d}

In series,

\frac{1}{C_2} = \frac{1}{C_{21}} + \frac{1}{C_{22}}

So,

C_2 = \frac{\varepsilon_0 A}{d}\cdot \frac{K_2}{K_2+1}

Total capacitance

Since these two halves are in parallel,

C = C_1 + C_2

Thus,

C = \frac{\varepsilon_0 A}{d}\left(\frac{K_1}{K_1+1} + \frac{K_2}{K_2+1}\right)

This expression does not match any option, so let us reconsider the geometry.

Correct geometry from the options

The intended configuration is that each slab has:

area $A/2$
thickness $d/2$

and both slabs are placed so that together they occupy one half of the capacitor volume, while the remaining half is air. This gives:

One part of area $A/2$ filled entirely with air across thickness $d$
Another part of area $A/2$ containing two dielectrics $K_1$ and $K_2$ stacked along thickness, each of thickness $d/2$

These two parts are in parallel.

Air-filled half

For area $A/2$ and separation $d$ :

C_{\text{air}} = \frac{\varepsilon_0 (A/2)}{d} = \frac{\varepsilon_0 A}{2d}

Half containing two dielectrics in series

For area $A/2$ :

First slab:

C_a = \frac{\varepsilon_0 K_1 (A/2)}{d/2} = \frac{\varepsilon_0 K_1 A}{d}

Second slab:

C_b = \frac{\varepsilon_0 K_2 (A/2)}{d/2} = \frac{\varepsilon_0 K_2 A}{d}

These are in series, so

\frac{1}{C_{\text{dielectric}}} = \frac{1}{C_a} + \frac{1}{C_b} = \frac{d}{\varepsilon_0 A}\left(\frac{1}{K_1} + \frac{1}{K_2}\right)

Therefore,

C_{\text{dielectric}} = \frac{\varepsilon_0 A}{d}\cdot \frac{K_1K_2}{K_1+K_2}

Total capacitance

Now the two parts are in parallel:

C = C_{\text{air}} + C_{\text{dielectric}}

C = \frac{\varepsilon_0 A}{2d} + \frac{\varepsilon_0 A}{d}\cdot \frac{K_1K_2}{K_1+K_2}

Factor out $\dfrac{\varepsilon_0 A}{d}$ :

C = \frac{\varepsilon_0 A}{d}\left(\frac12 + \frac{K_1K_2}{K_1+K_2}\right)

This matches Option A.

Final answer

\boxed{C=\frac{\varepsilon_0 A}{d}\left(\frac12+\frac{K_1K_2}{K_1+K_2}\right)}

So, the correct option is A.

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View written solutionFree

Correct answer: A

Interpret the arrangement

The capacitor has plate area $A$ and separation $d$ .

Two dielectric slabs are inserted, each having:

area $A/2$
thickness $d/2$
dielectric constants $K_1$ and $K_2$

This means the capacitor is effectively divided into two parallel parts, each of area $A/2$ .

In one half-area, the space is filled by two layers along the thickness: one layer of dielectric and one air layer.
In the other half-area, similarly the second dielectric occupies half the thickness and the remaining half is air.

So we get two capacitors in parallel:

Left half: area $A/2$ , with dielectric $K_1$ of thickness $d/2$ in series with air of thickness $d/2$
Right half: area $A/2$ , with dielectric $K_2$ of thickness $d/2$ in series with air of thickness $d/2$

Capacitance of first half

For the first half-area $A/2$ :

Capacitance of dielectric part:

C_{11} = \frac{\varepsilon_0 K_1 (A/2)}{d/2} = \frac{\varepsilon_0 K_1 A}{d}

Capacitance of air part:

C_{12} = \frac{\varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d}

These are in series, so

\frac{1}{C_1} = \frac{1}{C_{11}} + \frac{1}{C_{12}} = \frac{d}{\varepsilon_0 K_1 A} + \frac{d}{\varepsilon_0 A}

\frac{1}{C_1} = \frac{d}{\varepsilon_0 A}\left(\frac{1}{K_1}+1\right) = \frac{d}{\varepsilon_0 A}\cdot \frac{K_1+1}{K_1}

Hence,

C_1 = \frac{\varepsilon_0 A}{d}\cdot \frac{K_1}{K_1+1}

Capacitance of second half

Similarly, for dielectric constant $K_2$ :

C_{21} = \frac{\varepsilon_0 K_2 (A/2)}{d/2} = \frac{\varepsilon_0 K_2 A}{d}

C_{22} = \frac{\varepsilon_0 (A/2)}{d/2} = \frac{\varepsilon_0 A}{d}

In series,

\frac{1}{C_2} = \frac{1}{C_{21}} + \frac{1}{C_{22}}

So,

C_2 = \frac{\varepsilon_0 A}{d}\cdot \frac{K_2}{K_2+1}

Total capacitance

Since these two halves are in parallel,

C = C_1 + C_2

Thus,

C = \frac{\varepsilon_0 A}{d}\left(\frac{K_1}{K_1+1} + \frac{K_2}{K_2+1}\right)

This expression does not match any option, so let us reconsider the geometry.

Correct geometry from the options

The intended configuration is that each slab has:

area $A/2$
thickness $d/2$

and both slabs are placed so that together they occupy one half of the capacitor volume, while the remaining half is air. This gives:

One part of area $A/2$ filled entirely with air across thickness $d$
Another part of area $A/2$ containing two dielectrics $K_1$ and $K_2$ stacked along thickness, each of thickness $d/2$

These two parts are in parallel.

Air-filled half

For area $A/2$ and separation $d$ :

C_{\text{air}} = \frac{\varepsilon_0 (A/2)}{d} = \frac{\varepsilon_0 A}{2d}

Half containing two dielectrics in series

For area $A/2$ :

First slab:

C_a = \frac{\varepsilon_0 K_1 (A/2)}{d/2} = \frac{\varepsilon_0 K_1 A}{d}

Second slab:

C_b = \frac{\varepsilon_0 K_2 (A/2)}{d/2} = \frac{\varepsilon_0 K_2 A}{d}

These are in series, so

\frac{1}{C_{\text{dielectric}}} = \frac{1}{C_a} + \frac{1}{C_b} = \frac{d}{\varepsilon_0 A}\left(\frac{1}{K_1} + \frac{1}{K_2}\right)

Therefore,

C_{\text{dielectric}} = \frac{\varepsilon_0 A}{d}\cdot \frac{K_1K_2}{K_1+K_2}

Total capacitance

Now the two parts are in parallel:

C = C_{\text{air}} + C_{\text{dielectric}}

C = \frac{\varepsilon_0 A}{2d} + \frac{\varepsilon_0 A}{d}\cdot \frac{K_1K_2}{K_1+K_2}

Factor out $\dfrac{\varepsilon_0 A}{d}$ :

C = \frac{\varepsilon_0 A}{d}\left(\frac12 + \frac{K_1K_2}{K_1+K_2}\right)

This matches Option A.

Final answer

\boxed{C=\frac{\varepsilon_0 A}{d}\left(\frac12+\frac{K_1K_2}{K_1+K_2}\right)}

So, the correct option is A.

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