Question

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 $\ldots \ldots \ldots \in_0 J$. (Where $\in_0$ is the permittivity of free space)

Answer 1

$ 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$