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NVAMediumJEE 2023Capacitors & Dielectrics

JEE Physics 2023 Question with Solution

A capacitor has a capacitance of 5 μF5 \, \mu \text{F} when its parallel plates are separated by an air medium of thickness dd. A slab of material with a dielectric constant of 1.51.5, having an area equal to that of the plates but with thickness d/2d/2, is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be _____ μF\mu \text{F} :

Answer

Correct answer:6

Step-by-step solution

Standard Method

Given: Initial capacitance is 5 μF5 \, \mu \text{F} for a parallel plate capacitor with plate separation dd and air as dielectric.

Find: The new capacitance when a slab of dielectric constant K=1.5K = 1.5 and thickness d/2d/2 is inserted.

Parallel plate capacitor diagram showing plate area A, separation d, initial capacitance epsilon naught A by d equals 5 microfarad, and a dielectric slab of K equals 1.5 inserted over thickness d by 2 with remaining air gap d by 2.

The arrangement behaves like two capacitors in series along the separation: one region of dielectric thickness d/2d/2 and one region of air thickness d/2d/2.

So, the effective capacitance is

Cnew=ε0Ad/21.5+d/21C_{\text{new}} = \frac{\varepsilon_0 A}{\dfrac{d/2}{1.5} + \dfrac{d/2}{1}}

Using the initial capacitance,

ε0Ad=5 μF\frac{\varepsilon_0 A}{d} = 5 \, \mu \text{F}

Therefore,

Cnew=ε0Ad3+d2=ε0A5d6=65(ε0Ad)C_{\text{new}} = \frac{\varepsilon_0 A}{\dfrac{d}{3} + \dfrac{d}{2}} = \frac{\varepsilon_0 A}{\dfrac{5d}{6}} = \frac{6}{5}\left(\frac{\varepsilon_0 A}{d}\right)

Substituting ε0Ad=5 μF\dfrac{\varepsilon_0 A}{d} = 5 \, \mu \text{F},

Cnew=65×5 μF=6 μFC_{\text{new}} = \frac{6}{5} \times 5 \, \mu \text{F} = 6 \, \mu \text{F}

Therefore, the capacitance in the presence of the slab is 6 μF6 \, \mu \text{F}.

Series Combination Interpretation

Given: A dielectric slab fills only part of the plate separation, not the full gap.

Find: Why the effective gap changes in the denominator.

For a layered dielectric arrangement along the field direction, the separations add as effective thickness terms:

d1K1+d2K2\frac{d_1}{K_1} + \frac{d_2}{K_2}

Here,

d1=d2,K1=1.5,d2=d2,K2=1d_1 = \frac{d}{2}, \quad K_1 = 1.5, \quad d_2 = \frac{d}{2}, \quad K_2 = 1

Hence,

d/21.5+d/21=d3+d2=5d6\frac{d/2}{1.5} + \frac{d/2}{1} = \frac{d}{3} + \frac{d}{2} = \frac{5d}{6}

So the capacitor behaves like an air capacitor with effective separation 5d/65d/6, which is smaller than dd. Therefore the capacitance increases.

Since capacitance is inversely proportional to separation,

Cnew=d5d/6×5 μF=65×5 μF=6 μFC_{\text{new}} = \frac{d}{5d/6} \times 5 \, \mu \text{F} = \frac{6}{5} \times 5 \, \mu \text{F} = 6 \, \mu \text{F}

Therefore, the final answer is 66.

Common mistakes

  • Treating the slab and air regions as parallel combinations is incorrect because they are arranged along the field direction, so they act in series. Use effective thickness addition, not area addition.

  • Using K=1.5K = 1.5 as a multiplier of the whole capacitance is wrong because the dielectric does not fill the full separation. Apply the dielectric only to the thickness d/2d/2 occupied by the slab.

  • Forgetting that air has dielectric constant 11 leads to an incomplete denominator. The remaining half-gap must be included as d/21\dfrac{d/2}{1}.

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