Question 100 of 180

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

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Answer: 161

Here, Plate area$A=100{m}^{2}$
Using, Capacity of parallel plate capacitor is,
$C=\dfrac{k{\epsilon }_{0}A}{d}$; where $d$ is separation between the plates and $k$ is dielectric constant.
The capacitance with dielectric will be,
${C}_{1}=\dfrac{10{ϵ}_{0}(100)}{5}$
$=200{ϵ}_{0}$
The capacitance without dielectric or air will be,
${C}_{2}=\dfrac{{ϵ}_{0}(100)}{5}=20{ϵ}_{0}$
{C}_{1}&{C}_{2} are in series so,
${C}_{\text{eqv.}}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}$
$=\dfrac{4000{ϵ}_{0}}{220}$
$=160.9\times {10}^{-12}≃161pF$