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JEE Physics 2026 Question with Solution

A parallel plate capacitor has capacitance CC, when there is vacuum within the parallel plates. A sheet having thickness 13d\frac{1}{3}d of the separation between the plates and relative permittivity KK is introduced between the plates. The new capacitance of the system is:

  • A

    3KC2K+1\frac{3KC}{2K+1}

  • B

    CK2+K\frac{CK}{2+K}

  • C

    3CK2(2K+1)2\frac{3CK^2}{(2K+1)^2}

  • D

    4KC3K−1\frac{4KC}{3K-1}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The initial capacitance with vacuum is C=ϵ0AdC = \frac{\epsilon_0 A}{d}. A dielectric slab of thickness t=d3t = \frac{d}{3} and relative permittivity KK is inserted.

Find: The new equivalent capacitance of the system.

The arrangement can be treated as two capacitors in series:

  • dielectric part of thickness tt
  • remaining vacuum gap of thickness d−td-t

For the dielectric part,

C1=Kϵ0At=Kϵ0Ad/3=3Kϵ0AdC_1 = \frac{K\epsilon_0 A}{t} = \frac{K\epsilon_0 A}{d/3} = \frac{3K\epsilon_0 A}{d}

So, in terms of the original capacitance,

C1=3KCC_1 = 3KC

For the remaining air gap,

d−t=d−d3=2d3d-t = d-\frac{d}{3} = \frac{2d}{3}

Hence,

C2=ϵ0Ad−t=ϵ0A2d/3=3ϵ0A2dC_2 = \frac{\epsilon_0 A}{d-t} = \frac{\epsilon_0 A}{2d/3} = \frac{3\epsilon_0 A}{2d}

So,

C2=32CC_2 = \frac{3}{2}C

Since these two parts are in series,

1Ceq=1C1+1C2=13KC+1(3/2)C\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3KC} + \frac{1}{(3/2)C}

Therefore,

1Ceq=13KC+23C=1+2K3KC\frac{1}{C_{eq}} = \frac{1}{3KC} + \frac{2}{3C} = \frac{1+2K}{3KC}

On inverting,

Ceq=3KC1+2K=3KC2K+1C_{eq} = \frac{3KC}{1+2K} = \frac{3KC}{2K+1}

Therefore, the new capacitance is 3KC2K+1\frac{3KC}{2K+1} and the correct option is A.

Effective Distance Method

Given: A dielectric slab of thickness t=d3t = \frac{d}{3} is inserted between capacitor plates.

Find: The new capacitance.

Use the effective separation formula for a partially filled capacitor:

deff=(d−t)+tKd_{eff} = (d-t) + \frac{t}{K}

Substitute t=d3t = \frac{d}{3}:

deff=2d3+d3K=d(2K+1)3Kd_{eff} = \frac{2d}{3} + \frac{d}{3K} = \frac{d(2K+1)}{3K}

Now,

C′=ϵ0Adeff=ϵ0Ad⋅3K2K+1C' = \frac{\epsilon_0 A}{d_{eff}} = \frac{\epsilon_0 A}{d} \cdot \frac{3K}{2K+1}

Since C=ϵ0AdC = \frac{\epsilon_0 A}{d},

C′=3KC2K+1C' = \frac{3KC}{2K+1}

Therefore, the correct option is A.

Common mistakes

  • Treating the dielectric-filled part and air-gap part as capacitors in parallel is incorrect because the layers are stacked along the field direction. They are in series. Use series combination, not parallel addition.

  • Using the full plate separation dd as the dielectric thickness is wrong. The slab thickness is only d3\frac{d}{3}, so the remaining vacuum gap is 2d3\frac{2d}{3}.

  • Forgetting to rewrite C1C_1 and C2C_2 in terms of the original capacitance CC can lead to algebra mistakes. First use C=ϵ0AdC = \frac{\epsilon_0 A}{d}, then substitute to simplify the result.

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