In a parallel plate capacitor… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Electromagnetism

2021integermedium

In a parallel plate capacitor set up, the plate area of capacitor is $2{m}^{2}$ and the plates are separated by $1m$ . If the space between the plates are filled with a dielectric material of thickness $0.5m$ and are $2{m}^{2}$ (see figure) the capacitance of the set-up will be ${\epsilon }_{0}$ ________.

(Dielectric constant of the material $=3.2$ )

(Round off to the Nearest Integer)

JEE Main 2021 Physics, Capacitance — question figure

Official previous-year question

Held on 16 Mar 2021 · Verified 6 Jul 2026.

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Solution

The given system can be considered as a series combination of two capacitors. The equivalent capacitance can be given as,

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

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

$=\frac{2\times 2{\epsilon }_{0}}{\frac{1}{3.2}+1}=\frac{4\times 3.2}{4.2}{\epsilon }_{0}$

$=3.04{\epsilon }_{0}$

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Solution

The given system can be considered as a series combination of two capacitors. The equivalent capacitance can be given as,

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

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

$=\frac{2\times 2{\epsilon }_{0}}{\frac{1}{3.2}+1}=\frac{4\times 3.2}{4.2}{\epsilon }_{0}$

$=3.04{\epsilon }_{0}$

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Solution

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Solution

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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