![JEE Main 2020 Physics, Capacitance — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Capacitance/648b5a68417cc3fb48d66c5a/question_1__q_648b5a68417cc3fb48d66c5a__cdn-question-pool.getmarks.app__6d900902-6031-4cfe-ad66-57412f803ffd__63da0a95a7_final_ppt_sync.png) 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 $K(x)={K}_{0}(1+\alpha x)$ where $x$ is the distance measured from one of the plates. If $(\alpha d)<<1,$ the total capacitance of the system is best given by the expression:

JEE Main 2020 — Physics Electromagnetism

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JEE Main 2020 Physics, Capacitance — question figure

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 $K(x)={K}_{0}(1+\alpha x)$ where $x$ is the distance measured from one of the plates. If $(\alpha d)<<1,$ the total capacitance of the system is best given by the expression:

Official previous-year question

Held on 7 Jan 2020 · Verified 6 Jul 2026.

Options

A
$\frac{A{K}_{0}{\epsilon }_{0}}{d}(1+\frac{\alpha d}{2})$
B
$\frac{A{K}_{0}{\epsilon }_{0}}{d}[1+{(\frac{\alpha d}{2})}^{2}]$
C
$\frac{A{K}_{0}{\epsilon }_{0}}{d}(1+\frac{{\alpha }^{2}{d}^{2}}{2})$
D
$\frac{A{K}_{0}{\epsilon }_{0}}{d}(1+\alpha d)$

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Solution

Capacitance of element, $dC=\frac{{K}_{0}(1+\alpha x){\epsilon }_{0}A}{dx}$

$\int \frac{1}{dC}={\int }_{0}^{d}\frac{dx}{{K}_{0}{\epsilon }_{0}A(1+\alpha x)}$

$\Rightarrow \frac{1}{C}=\frac{1}{{K}_{0}{\epsilon }_{0}A\alpha }\mathrm{ln}(1+\alpha d)$

Given $\alpha d\ll 1$

$\Rightarrow \frac{1}{C}=\frac{1}{{K}_{0}{\epsilon }_{0}A\alpha }(\alpha d-\frac{{\alpha }^{2}{d}^{2}}{2})$

$\Rightarrow \frac{1}{C}=\frac{d}{{K}_{0}{\epsilon }_{0}A}(1-\frac{\alpha d}{2})$

$\Rightarrow C=\frac{{K}_{0}{\epsilon }_{0}A}{d}(1+\frac{\alpha d}{2})$

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Options

$\frac{A{K}_{0}{\epsilon }_{0}}{d}(1+\frac{\alpha d}{2})$

$\frac{A{K}_{0}{\epsilon }_{0}}{d}[1+{(\frac{\alpha d}{2})}^{2}]$

$\frac{A{K}_{0}{\epsilon }_{0}}{d}(1+\frac{{\alpha }^{2}{d}^{2}}{2})$

$\frac{A{K}_{0}{\epsilon }_{0}}{d}(1+\alpha d)$

Solution

Capacitance of element, $dC=\frac{{K}_{0}(1+\alpha x){\epsilon }_{0}A}{dx}$

$\int \frac{1}{dC}={\int }_{0}^{d}\frac{dx}{{K}_{0}{\epsilon }_{0}A(1+\alpha x)}$

$\Rightarrow \frac{1}{C}=\frac{1}{{K}_{0}{\epsilon }_{0}A\alpha }\mathrm{ln}(1+\alpha d)$

Given $\alpha d\ll 1$

$\Rightarrow \frac{1}{C}=\frac{1}{{K}_{0}{\epsilon }_{0}A\alpha }(\alpha d-\frac{{\alpha }^{2}{d}^{2}}{2})$

$\Rightarrow \frac{1}{C}=\frac{d}{{K}_{0}{\epsilon }_{0}A}(1-\frac{\alpha d}{2})$

$\Rightarrow C=\frac{{K}_{0}{\epsilon }_{0}A}{d}(1+\frac{\alpha d}{2})$

JEE Main 2020 — Physics Electromagnetism

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JEE Main 2020 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism