A parallel plate capacitor with width 4,cm , length 8,cm and separation between plates of 4,mm is co

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2. Electric Potential and Capacitance

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A parallel plate capacitor with width $4\,cm$, length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. A dielectric slab of dielectric constant $5$ having length $1\,cm$, width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be......... $\in_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)

A

$240$

B

$241$

C

$242$

D

$243$

(JEE MAIN-2022)

Solution

$C _{\text {eff }}=\left[\frac{\varepsilon_{0}(7 \times 4)}{4 / 10}+\frac{5 \varepsilon_{0}(1 \times 4)}{4 / 10}\right] \times 10^{-2}$

$C _{\text {eff }}=1.2 \varepsilon_{0}$

Energy $=\frac{1}{2} C _{\text {eff }} V ^{2}$

$=\frac{1}{2}(1.2) \varepsilon_{0}(20)(20)=240 \varepsilon_{0}$

Standard 12

Physics

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