A parallel plate capacitor has plate area 100, m ^{2} and plate separation of 10, m . space between

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2. Electric Potential and Capacitance

hard

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 $\varepsilon_{0}=8.85 \times 10^{-12} F \cdot m ^{-1}$ The value of $'x'$ to the nearest integer is............

A

$144$

B

$161$

C

$169$

D

$152$

(JEE MAIN-2021)

Solution

$A =100 m ^{2}$

Using $C =\frac{ k \in_{0} A }{ d }$

$C _{1}=\frac{10 \epsilon_{0}(100)}{5}$

$=200 \in_{0}$

$C _{2}=\frac{\epsilon_{0}(100)}{5}=20 \epsilon_{0}$

$C _{1} \& C _{2}$ are in series so $C _{ eqv. }=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}$

Standard 12

Physics

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