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Capacitor question

2019 · Shift 1 · Q49
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Capacitor question

2019 · Shift 1 · Q49

JEE AdvancedPhysicsCapacitorNumerical+3 / −1
A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness, δ=dN\delta = {d \over N}δ=Nd​. The dielectric constant of the mth layer is Km=K(1+mN){K_m} = K\left( {1 + {m \over N}} \right)Km​=K(1+Nm​). For a very large N(>103), the capacitance C is α(Kε0Adln⁡2)\alpha \left( {{{K{\varepsilon _0}A} \over {d\ln 2}}} \right)α(dln2Kε0​A​) The value of α\alphaα will be .................. [∈\in∈ 0 is the permittivity of free space.]
Numerical answer
View written solutionFree

Correct answer: 1

  1. Model the layered dielectric as capacitors in series

Since the dielectric layers are parallel to the plates, the electric field passes through them one after another along the separation direction. Hence the layers behave like series capacitors.

For the mmmth layer:

  • thickness δ=dN\delta = \dfrac{d}{N}δ=Nd​
  • dielectric constant Km=K(1+mN)K_m = K\left(1+\dfrac{m}{N}\right)Km​=K(1+Nm​)
  • permittivity εm=Kmε0\varepsilon_m = K_m \varepsilon_0εm​=Km​ε0​

So its capacitance is

Cm=εmAδ=K(1+mN)ε0Ad/N=NKε0Ad(1+mN).C_m = \frac{\varepsilon_m A}{\delta} = \frac{K\left(1+\frac{m}{N}\right)\varepsilon_0 A}{d/N} = \frac{N K\varepsilon_0 A}{d}\left(1+\frac{m}{N}\right).Cm​=δεm​A​=d/NK(1+Nm​)ε0​A​=dNKε0​A​(1+Nm​).
  1. Use the series combination formula

For capacitors in series,

1C=∑m=1N1Cm.\frac{1}{C} = \sum_{m=1}^{N} \frac{1}{C_m}.C1​=m=1∑N​Cm​1​.

Thus,

1C=∑m=1NdNKε0A⋅11+m/N=dNKε0A∑m=1N11+m/N.\frac{1}{C} = \sum_{m=1}^{N} \frac{d}{N K\varepsilon_0 A}\cdot \frac{1}{1+m/N} = \frac{d}{N K\varepsilon_0 A} \sum_{m=1}^{N} \frac{1}{1+m/N}.C1​=m=1∑N​NKε0​Ad​⋅1+m/N1​=NKε0​Ad​m=1∑N​1+m/N1​.

Now simplify the summand:

11+m/N=NN+m.\frac{1}{1+m/N} = \frac{N}{N+m}.1+m/N1​=N+mN​.

Therefore,

1C=dNKε0A∑m=1NNN+m=dKε0A∑m=1N1N+m.\frac{1}{C} = \frac{d}{N K\varepsilon_0 A} \sum_{m=1}^{N} \frac{N}{N+m} = \frac{d}{K\varepsilon_0 A} \sum_{m=1}^{N} \frac{1}{N+m}.C1​=NKε0​Ad​m=1∑N​N+mN​=Kε0​Ad​m=1∑N​N+m1​.

So,

1C=dKε0A∑r=N+12N1r.\frac{1}{C} = \frac{d}{K\varepsilon_0 A} \sum_{r=N+1}^{2N} \frac{1}{r}.C1​=Kε0​Ad​r=N+1∑2N​r1​.
  1. Approximate the sum for very large NNN

For large NNN,

∑r=N+12N1r≈∫N2Ndxx=ln⁡(2N)−ln⁡N=ln⁡2.\sum_{r=N+1}^{2N} \frac{1}{r} \approx \int_N^{2N} \frac{dx}{x} = \ln(2N)-\ln N = \ln 2.r=N+1∑2N​r1​≈∫N2N​xdx​=ln(2N)−lnN=ln2.

Hence,

1C≈dKε0Aln⁡2.\frac{1}{C} \approx \frac{d}{K\varepsilon_0 A}\ln 2.C1​≈Kε0​Ad​ln2.

So,

C≈Kε0Adln⁡2.C \approx \frac{K\varepsilon_0 A}{d\ln 2}.C≈dln2Kε0​A​.
  1. Compare with the given form

Given,

C=α(Kε0Adln⁡2).C = \alpha \left(\frac{K\varepsilon_0 A}{d\ln 2}\right).C=α(dln2Kε0​A​).

Comparing, we get

α=1.\alpha = 1.α=1.
  1. Verification with stored answer

Stored correct answer: 111

Our derived answer: 111

They match.

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