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 ______ $\varepsilon_0$ J. (Where $\varepsilon_0$ is the permittivity of free space)
Numerical value type. Enter your answer.
Answer: 240
The slab fills the gap over $1\times4\ \text{cm}^2$; the remaining $7\times4\ \text{cm}^2$ is air. These act in parallel:
$$C = \frac{\varepsilon_0}{d}\left(A_{\text{air}} + KA_{\text{slab}}\right) = \frac{\varepsilon_0}{4\times10^{-3}}\left(28 + 5\times4\right)\times10^{-4} = 1.2\,\varepsilon_0$$
$$U = \frac12CV^2 = \frac12(1.2\,\varepsilon_0)(20)^2 = 240\,\varepsilon_0\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics