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JEE Main 20207 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Morning Shift
Subject: Physics
Chapter: Capacitance

JEE Main 2020 · 7 January · Morning Shift · Physics · Capacitance

A parallel plate capacitor has plates of area A separated by...

A parallel plate capacitor has plates of area A separated by distance d between them. It is filled with a dielectric

A parallel plate capacitor has plates of area A separated by distance

d

between them. It is filled with a dielectric which has a dielectric constant that varies as $

K\left(x\right)

= {K}_{0}

\left(1 + \alphax\right)

where

x

is the distance measured from one of the plates. If

\left(\alphad\right)

< < 1,$ the total capacitance of the system is best given by the expression:

Options

A $\dfrac{A{K}_{0}{ε}_{0}}{d}\left(1 + \dfrac{\alphad}{2}\right)$
B $\dfrac{A{K}_{0}{ε}_{0}}{d}\left[1 + {\left(\dfrac{\alphad}{2}\right)}^{2}\right]$
C $\dfrac{A{K}_{0}{ε}_{0}}{d}\left(1 + \dfrac{{\alpha}^{2}{d}^{2}}{2}\right)$
D $\dfrac{A{K}_{0}{ε}_{0}}{d}\left(1 + \alphad\right)$

Correct answer

A. $\dfrac{A{K}_{0}{ε}_{0}}{d}\left(1 + \dfrac{\alphad}{2}\right)$

Step-by-step solution

Capacitance of element,

d C = \dfrac{{K}_{0}\left(1 + \alphax\right){ε}_{0}A}{dx}

∫ \dfrac{1}{d C} = {∫}_{0}^{d}\dfrac{dx}{{K}_{0}{ε}_{0}A\left(1 + \alphax\right)}

⇒ \dfrac{1}{C} = \dfrac{1}{{K}_{0}{ε}_{0}A\alpha}\ln \left(1 + \alphad\right)

Given

\alphad ≪ 1

⇒ \dfrac{1}{C} = \dfrac{1}{{K}_{0}{ε}_{0}A\alpha}\left(\alphad - \dfrac{{\alpha}^{2}{d}^{2}}{2}\right)

⇒ \dfrac{1}{C} = \dfrac{d}{{K}_{0}{ε}_{0}A}\left(1 - \dfrac{\alphad}{2}\right)

⇒ C = \dfrac{{K}_{0}{ε}_{0}A}{d}\left(1 + \dfrac{\alphad}{2}\right)

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