Question

A parallel plate capacitor with width $4cm$, length $8cm$ and separation between the plates of $4mm$ is connected to a battery of $20V$. A dielectric slab of dielectric constant $5$ having length $1cm$, width $4cm$ and thickness $4mm$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be $\dots \dots \dots {\in }_{0}J$. (Where ${\in }_0$ is the permittivity of free space)

Answer 1

Answer: 240

$C_{\text{eff }}=\left[\frac{{\epsilon }_0(7\times 4)}{4/10}+\frac{5{\epsilon }_0(1\times 4)}{4/10}\right]\times 10^{-2}$
$C_{\text{eff }}=1.2{\epsilon }_0$
Energy $=\frac{1}{2}{C}_{\text{eff }}{V}^2$
$=\frac{1}{2}(1.2){\epsilon }_0(20)(20)=240{\epsilon }_0$