JEE Main 2021 — Capacitance Question with Solution

Here, Plate area$A=100{ \mathrm{m}}^2$
Using, Capacity of parallel plate capacitor is,

$C=\frac{k{\epsilon }_0 A}{ d}$; where $d$ is separation between the plates and $k$ is dielectric constant.

The capacitance with dielectric will be,
$C_1=\frac{10{ϵ}_0\left(100\right)}{5}$
$=200{ϵ}_0$

The capacitance without dielectric or air will be,
$C_2=\frac{{ϵ}_0\left(100\right)}{5}=20{ϵ}_0$
$C_1 \& C_2$ are in series so,

 $C_{\text{eqv. }}=\frac{C_1{C}_2}{C_1+C_2}$

$=\frac{4000{ϵ}_0}{220}$
$=160.9\times {10}^{-12}\simeq 161\mathrm{pF}$