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JEE Main 201419 AprilOnline
Exam
JEE Main
Year
2014
Date
19 April
Mode
Online
Subject
Physics
Chapter
Capacitance

JEE Main 2014 · 19 April · Online · Physics · Capacitance

The gap between the plates of a parallel plate capacitor of area A...

The gap between the plates of a parallel plate capacitor of area AA and distance between plates dd, is filled with a dielectric whose relative permittivity varies linearly from ϵ1{ϵ}_{1} at one plate to ϵ2{ϵ}_{2} at the other. The capacitance of the capacitor isThe gap between the plates of a parallel plate capacitor of area A and distance between plates d , is filled with a

Options

  1. Aϵ0(∈2−∈1)A[dln(∈2/∈1)]\dfrac{{ϵ}_{0}\left({\in}_{2} - {\in}_{1}\right)A}{\left[dln\left({\in}_{2} / {\in}_{1}\right)\right]}
  2. Bϵ0(∈2+∈1)A2d\dfrac{{ϵ}_{0}\left({\in}_{2} + {\in}_{1}\right)A}{2d}
  3. Cϵ0(∈1+∈2)Ad\dfrac{{ϵ}_{0}\left({\in}_{1} + {\in}_{2}\right)A}{d}
  4. Dϵ0A[dln⁡(∈2/∈1)]\dfrac{{ϵ}_{0}A}{\left[d \ln \left({\in}_{2} / {\in}_{1}\right)\right]}

Correct answer

A. ϵ0(∈2−∈1)A[dln(∈2/∈1)]\dfrac{{ϵ}_{0}\left({\in}_{2} - {\in}_{1}\right)A}{\left[dln\left({\in}_{2} / {\in}_{1}\right)\right]}

Step-by-step solution

1dC=dxϵ0ϵA\dfrac{1}{dC} = \dfrac{dx}{{ϵ}_{0}ϵA}

=dxϵ0[ϵ1+(ϵ2−ϵ1)dx]A= \dfrac{dx}{{ϵ}_{0}\left[{ϵ}_{1} + \dfrac{\left({ϵ}_{2} - {ϵ}_{1}\right)}{d}x\right]A}

1C=∫1dC=1∈0A∫0xdx∈1+(∈2−∈1d)\dfrac{1}{C} = ∫ \dfrac{1}{dC} = \dfrac{1}{{\in}_{0}A}\mathop{∫}_{0}^{x}\dfrac{dx}{{\in}_{1} + \left(\dfrac{{\in}_{2} - {\in}_{1}}{d}\right)}

=ln⁡(∈2/∈1)d∈0A(∈2−∈1)= \dfrac{\ln \left({\in}_{2} / {\in}_{1}\right)d}{{\in}_{0}A\left({\in}_{2} - {\in}_{1}\right)}

C=∈0A(∈2−∈1)ln⁡(∈2/∈1)⋅dC = \dfrac{{\in}_{0}A\left({\in}_{2} - {\in}_{1}\right)}{\ln \left({\in}_{2} / {\in}_{1}\right) \cdot d}

The gap between the plates of a parallel plate capacitor of area A and distance between plates d , is filled with a

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