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Q. A parallel plate capacitor has plate area $100\,m ^{2}$ and plate separation of $10\,m$. The space between the plates is filled up to a thickness 5 $m$ with a material of dielectric constant of $10 .$ The resultant capacitance of the system is 'x' $pF$. The value of $\varepsilon_{0}=8.85 \times 10^{-12} F.m ^{-1}$.The value of ' $x$ ' to the nearest integer is ______.

JEE MainJEE Main 2021Electrostatic Potential and Capacitance

Solution:

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$A =100\,m ^{2}$
Using $C =\frac{ k \in_{0} A }{ d }$
$C _{1}=\frac{10 \in_{0}(100)}{5}$
$=200 \in_{0}$
$C _{2}=\frac{\epsilon_{0}(100)}{5}=20 \in_{0}$
$C _{1} \& C _{2}$ are in series so $C _{\text {eqv. }}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}$
$=\frac{4000 \in_{0}}{220}$
$=160.9 \times 10^{-12} \simeq 161 pF$